∫IntegralsसमाकलनIntegrals

1. (i) We look for a function whose derivative is cos 2x; sin 2x gives 2 cos 2x, so halve it.(i) ऐसा फलन खोजते हैं जिसका अवकलज cos 2x हो; sin 2x से 2 cos 2x मिलता है, इसलिए आधा कर दिया।(i) Aisa function dhoondhte hain jiska derivative cos 2x ho; sin 2x se 2 cos 2x milta hai, isliye aadha kar diya.
   \dfrac{d}{dx}(\sin 2x) = 2\cos 2x \Rightarrow \cos 2x = \dfrac{d}{dx}\left(\dfrac{1}{2}\sin 2x\right)
2. (ii) The derivative of x³ is 3x² and of x⁴ is 4x³.(ii) x³ का अवकलज 3x² और x⁴ का 4x³ है।(ii) x³ ka derivative 3x² aur x⁴ ka 4x³ hai.
   \dfrac{d}{dx}(x^3 + x^4) = 3x^2 + 4x^3
3. (iii) Both cases combine into log |x|.(iii) दोनों स्थितियाँ मिलकर log |x| बनती हैं।(iii) Dono cases milkar log |x| bante hain.
   \dfrac{d}{dx}(\log x) = \dfrac{1}{x}\ (x > 0),\quad \dfrac{d}{dx}[\log(-x)] = \dfrac{1}{x}\ (x < 0)

✔ (i) \dfrac{1}{2}\sin 2x (ii) x^3 + x^4 (iii) \log |x| — each "+ C" also works.(i) \dfrac{1}{2}\sin 2x (ii) x^3 + x^4 (iii) \log |x| — हर एक में "+ C" भी सही है।(i) \dfrac{1}{2}\sin 2x (ii) x^3 + x^4 (iii) \log |x| — har ek mein "+ C" bhi sahi hai.

1. (i) Split the fraction, then use the power rule on each term.(i) भिन्न को तोड़िए, फिर हर पद पर घात का नियम लगाइए।(i) Fraction ko todiye, phir har term par power rule lagaiye.
   \displaystyle\int \frac{x^3 - 1}{x^2}\,dx = \int x\,dx - \int x^{-2}\,dx = \frac{x^2}{2} - \frac{x^{-1}}{-1} + C = \frac{x^2}{2} + \frac{1}{x} + C
2. (ii) Add 1 to the power 2/3 to get 5/3 and divide by 5/3.(ii) घात 2/3 में 1 जोड़कर 5/3 बनाइए और 5/3 से भाग दीजिए।(ii) Power 2/3 mein 1 jodkar 5/3 banaiye aur 5/3 se divide kijiye.
   \displaystyle\int (x^{2/3} + 1)\,dx = \frac{x^{5/3}}{5/3} + x + C = \frac{3}{5}x^{5/3} + x + C
3. (iii) Integrate term by term (property V).(iii) पद-दर-पद समाकलन कीजिए (गुणधर्म V)।(iii) Term-by-term integrate kijiye (property V).
   \displaystyle\int x^{3/2}dx + 2\int e^x dx - \int \frac{1}{x}dx = \frac{2}{5}x^{5/2} + 2e^x - \log |x| + C

✔ (i) \dfrac{x^2}{2} + \dfrac{1}{x} + C (ii) \dfrac{3}{5}x^{5/3} + x + C (iii) \dfrac{2}{5}x^{5/2} + 2e^x - \log |x| + C(i) \dfrac{x^2}{2} + \dfrac{1}{x} + C (ii) \dfrac{3}{5}x^{5/3} + x + C (iii) \dfrac{2}{5}x^{5/2} + 2e^x - \log |x| + C(i) \dfrac{x^2}{2} + \dfrac{1}{x} + C (ii) \dfrac{3}{5}x^{5/3} + x + C (iii) \dfrac{2}{5}x^{5/2} + 2e^x - \log |x| + C

1. (i) Two standard formulae.(i) दो मानक सूत्र।(i) Do standard formulae.
   \displaystyle\int \sin x\,dx + \int \cos x\,dx = -\cos x + \sin x + C
2. (ii) Open the bracket; both pieces are in the table.(ii) कोष्ठक खोलिए; दोनों भाग सारणी में हैं।(ii) Bracket kholiye; dono pieces table mein hain.
   \displaystyle\int \operatorname{cosec}^2 x\,dx + \int \operatorname{cosec} x\cot x\,dx = -\cot x - \operatorname{cosec} x + C
3. (iii) Split, then write sin x/cos² x as tan x · sec x.(iii) तोड़िए, फिर sin x/cos² x को tan x · sec x लिखिए।(iii) Todiye, phir sin x/cos² x ko tan x · sec x likhiye.
   \displaystyle\int \frac{1}{\cos^2 x}dx - \int \frac{\sin x}{\cos^2 x}dx = \int \sec^2 x\,dx - \int \tan x\sec x\,dx = \tan x - \sec x + C

✔ (i) -\cos x + \sin x + C (ii) -\cot x - \operatorname{cosec} x + C (iii) \tan x - \sec x + C(i) -\cos x + \sin x + C (ii) -\cot x - \operatorname{cosec} x + C (iii) \tan x - \sec x + C(i) -\cos x + \sin x + C (ii) -\cot x - \operatorname{cosec} x + C (iii) \tan x - \sec x + C

1. One anti derivative is x⁴ − 6x; the general one adds C.एक प्रतिअवकलज x⁴ − 6x है; व्यापक में C जोड़ते हैं।Ek anti derivative x⁴ − 6x hai; general mein C jodte hain.
   \dfrac{d}{dx}(x^4 - 6x) = 4x^3 - 6 \Rightarrow F(x) = x^4 - 6x + C
2. The extra condition fixes one member of the family.अतिरिक्त प्रतिबंध से कुल का एक ही सदस्य तय हो जाता है।Extra condition se family ka ek hi member fix ho jaata hai.
   F(0) = 0 - 0 + C = 3 \Rightarrow C = 3

✔ F(x) = x^4 - 6x + 3F(x) = x^4 - 6x + 3F(x) = x^4 - 6x + 3

1. (i) The derivative of mx is m.(i) mx का अवकलज m है।(i) mx ka derivative m hai.
   mx = t \Rightarrow dx = \dfrac{dt}{m};\ \displaystyle\int \sin mx\,dx = \frac{1}{m}\int \sin t\,dt = -\frac{1}{m}\cos mx + C
2. (ii) 2x is exactly the derivative of x² + 1.(ii) 2x ठीक x² + 1 का अवकलज है।(ii) 2x bilkul x² + 1 ka derivative hai.
   x^2 + 1 = t \Rightarrow 2x\,dx = dt;\ \displaystyle\int 2x\sin(x^2+1)\,dx = \int \sin t\,dt = -\cos(x^2 + 1) + C
3. (iii) dx/√x becomes 2 dt.(iii) dx/√x बन जाता है 2 dt।(iii) dx/√x ban jaata hai 2 dt.
   \sqrt{x} = t \Rightarrow \dfrac{dx}{2\sqrt{x}} = dt;\ \displaystyle\int 2\tan^4 t\,\sec^2 t\,dt
4. (iii) A second substitution, then the power rule.(iii) दूसरा प्रतिस्थापन, फिर घात का नियम।(iii) Doosra substitution, phir power rule.
   \tan t = u \Rightarrow \sec^2 t\,dt = du;\ \displaystyle 2\int u^4\,du = \frac{2u^5}{5} + C = \frac{2}{5}\tan^5\sqrt{x} + C
5. (iv) 1/(1 + x²) is the derivative of tan⁻¹x.(iv) 1/(1 + x²), tan⁻¹x का अवकलज है।(iv) 1/(1 + x²), tan⁻¹x ka derivative hai.
   \tan^{-1}x = t \Rightarrow \dfrac{dx}{1 + x^2} = dt;\ \displaystyle\int \sin t\,dt = -\cos(\tan^{-1}x) + C

✔ (i) -\dfrac{1}{m}\cos mx + C (ii) -\cos(x^2 + 1) + C (iii) \dfrac{2}{5}\tan^5\sqrt{x} + C (iv) -\cos(\tan^{-1}x) + C(i) -\dfrac{1}{m}\cos mx + C (ii) -\cos(x^2 + 1) + C (iii) \dfrac{2}{5}\tan^5\sqrt{x} + C (iv) -\cos(\tan^{-1}x) + C(i) -\dfrac{1}{m}\cos mx + C (ii) -\cos(x^2 + 1) + C (iii) \dfrac{2}{5}\tan^5\sqrt{x} + C (iv) -\cos(\tan^{-1}x) + C

1. (i) Keep one sin x for dt; turn sin² x into 1 − cos² x.(i) dt के लिए एक sin x रखिए; sin² x को 1 − cos² x बनाइए।(i) dt ke liye ek sin x rakhiye; sin² x ko 1 − cos² x banaiye.
   \displaystyle\int \sin^2 x\cos^2 x\,\sin x\,dx = \int (1 - \cos^2 x)\cos^2 x\,\sin x\,dx
2. (i) Now it is a polynomial in t.(i) अब यह t में एक बहुपद है।(i) Ab ye t mein ek polynomial hai.
   \cos x = t,\ -\sin x\,dx = dt:\ \displaystyle -\int (t^2 - t^4)\,dt = -\frac{t^3}{3} + \frac{t^5}{5} + C = -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C
3. (ii) Expand sin(t − a).(ii) sin(t − a) को फैलाइए।(ii) sin(t − a) ko expand kijiye.
   x + a = t,\ dx = dt:\ \displaystyle\int \frac{\sin(t - a)}{\sin t}\,dt = \int \frac{\sin t\cos a - \cos t\sin a}{\sin t}\,dt = \cos a\int dt - \sin a\int \cot t\,dt
4. (ii) a cos a is a constant, so it joins C.(ii) a cos a एक अचर है, इसलिए वह C में मिल जाता है।(ii) a cos a ek constant hai, isliye wo C mein mil jaata hai.
   = (x + a)\cos a - \sin a\log|\sin(x + a)| + C_1 = x\cos a - \sin a\log|\sin(x + a)| + C
5. (iii) Write tan x = sin x/cos x, then split the numerator cleverly.(iii) tan x = sin x/cos x लिखिए, फिर अंश को चतुराई से तोड़िए।(iii) tan x = sin x/cos x likhiye, phir numerator ko chalaaki se todiye.
   \displaystyle\int \frac{\cos x\,dx}{\cos x + \sin x} = \frac{1}{2}\int \frac{(\cos x + \sin x) + (\cos x - \sin x)}{\cos x + \sin x}\,dx
6. (iii) cos x − sin x is the derivative of cos x + sin x (put t = cos x + sin x).(iii) cos x − sin x, cos x + sin x का अवकलज है (t = cos x + sin x रखिए)।(iii) cos x − sin x, cos x + sin x ka derivative hai (t = cos x + sin x rakhiye).
   = \dfrac{x}{2} + \dfrac{1}{2}\displaystyle\int \frac{\cos x - \sin x}{\cos x + \sin x}\,dx = \frac{x}{2} + \frac{1}{2}\log|\cos x + \sin x| + C

✔ (i) -\dfrac{\cos^3 x}{3} + \dfrac{\cos^5 x}{5} + C (ii) x\cos a - \sin a\log|\sin(x + a)| + C (iii) \dfrac{x}{2} + \dfrac{1}{2}\log|\cos x + \sin x| + C(i) -\dfrac{\cos^3 x}{3} + \dfrac{\cos^5 x}{5} + C (ii) x\cos a - \sin a\log|\sin(x + a)| + C (iii) \dfrac{x}{2} + \dfrac{1}{2}\log|\cos x + \sin x| + C(i) -\dfrac{\cos^3 x}{3} + \dfrac{\cos^5 x}{5} + C (ii) x\cos a - \sin a\log|\sin(x + a)| + C (iii) \dfrac{x}{2} + \dfrac{1}{2}\log|\cos x + \sin x| + C

1. (i) cos 2x = 2cos² x − 1.(i) cos 2x = 2cos² x − 1।(i) cos 2x = 2cos² x − 1.
   \displaystyle\int \cos^2 x\,dx = \int \frac{1 + \cos 2x}{2}\,dx = \frac{x}{2} + \frac{1}{4}\sin 2x + C
2. (ii) 2 sin A cos B = sin(A + B) + sin(A − B) with A = 2x, B = 3x; sin(−x) = −sin x.(ii) A = 2x, B = 3x लेकर 2 sin A cos B = sin(A + B) + sin(A − B); sin(−x) = −sin x।(ii) A = 2x, B = 3x lekar 2 sin A cos B = sin(A + B) + sin(A − B); sin(−x) = −sin x.
   \sin 2x\cos 3x = \dfrac{1}{2}[\sin 5x - \sin x]
3. (ii) Integrate each sine.(ii) हर sine का समाकलन कीजिए।(ii) Har sine ko integrate kijiye.
   \displaystyle\frac{1}{2}\int (\sin 5x - \sin x)\,dx = -\frac{1}{10}\cos 5x + \frac{1}{2}\cos x + C
4. (iii) sin 3x = 3 sin x − 4 sin³ x.(iii) sin 3x = 3 sin x − 4 sin³ x।(iii) sin 3x = 3 sin x − 4 sin³ x.
   \displaystyle\int \sin^3 x\,dx = \frac{3}{4}\int \sin x\,dx - \frac{1}{4}\int \sin 3x\,dx = -\frac{3}{4}\cos x + \frac{1}{12}\cos 3x + C
5. (iii) NCERT's second way: put cos x = t. Both answers differ only by a constant.(iii) NCERT का दूसरा तरीका: cos x = t रखिए। दोनों उत्तरों में केवल एक अचर का अंतर है।(iii) NCERT ka doosra tareeka: cos x = t rakhiye. Dono answers mein sirf ek constant ka farak hai.
   \text{Alternatively: } \displaystyle\int (1 - \cos^2 x)\sin x\,dx = -\cos x + \frac{\cos^3 x}{3} + C

✔ (i) \dfrac{x}{2} + \dfrac{1}{4}\sin 2x + C (ii) -\dfrac{1}{10}\cos 5x + \dfrac{1}{2}\cos x + C (iii) -\dfrac{3}{4}\cos x + \dfrac{1}{12}\cos 3x + C(i) \dfrac{x}{2} + \dfrac{1}{4}\sin 2x + C (ii) -\dfrac{1}{10}\cos 5x + \dfrac{1}{2}\cos x + C (iii) -\dfrac{3}{4}\cos x + \dfrac{1}{12}\cos 3x + C(i) \dfrac{x}{2} + \dfrac{1}{4}\sin 2x + C (ii) -\dfrac{1}{10}\cos 5x + \dfrac{1}{2}\cos x + C (iii) -\dfrac{3}{4}\cos x + \dfrac{1}{12}\cos 3x + C

1. (i) Formula (1) with a = 4.(i) a = 4 लेकर सूत्र (1)।(i) a = 4 lekar formula (1).
   \displaystyle\int \frac{dx}{x^2 - 4^2} = \frac{1}{8}\log\left|\frac{x - 4}{x + 4}\right| + C
2. (ii) Complete the square, then formula (5) with a = 1.(ii) पूर्ण वर्ग बनाइए, फिर a = 1 लेकर सूत्र (5)।(ii) Perfect square banaiye, phir a = 1 lekar formula (5).
   2x - x^2 = 1 - (x - 1)^2;\ x - 1 = t:\ \displaystyle\int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}t + C = \sin^{-1}(x - 1) + C

✔ (i) \dfrac{1}{8}\log\left|\dfrac{x - 4}{x + 4}\right| + C (ii) \sin^{-1}(x - 1) + C(i) \dfrac{1}{8}\log\left|\dfrac{x - 4}{x + 4}\right| + C (ii) \sin^{-1}(x - 1) + C(i) \dfrac{1}{8}\log\left|\dfrac{x - 4}{x + 4}\right| + C (ii) \sin^{-1}(x - 1) + C

1. (i) Complete the square; formula (3) with t = x − 3, a = 2.(i) पूर्ण वर्ग; t = x − 3, a = 2 लेकर सूत्र (3)।(i) Perfect square; t = x − 3, a = 2 lekar formula (3).
   x^2 - 6x + 13 = (x - 3)^2 + 2^2 \Rightarrow \displaystyle\int \frac{dx}{(x - 3)^2 + 2^2} = \frac{1}{2}\tan^{-1}\frac{x - 3}{2} + C
2. (ii) Take 3 out and complete the square: (13/6)² + 10/3 = 289/36 = (17/6)².(ii) 3 बाहर निकालकर पूर्ण वर्ग: (13/6)² + 10/3 = 289/36 = (17/6)²।(ii) 3 bahar nikaalkar perfect square: (13/6)² + 10/3 = 289/36 = (17/6)².
   3x^2 + 13x - 10 = 3\left[\left(x + \dfrac{13}{6}\right)^2 - \left(\dfrac{17}{6}\right)^2\right]
3. (ii) Formula (1); the constant (1/17) log(1/3) is absorbed into C.(ii) सूत्र (1); अचर (1/17) log(1/3) को C में मिला दिया।(ii) Formula (1); constant (1/17) log(1/3) ko C mein mila diya.
   \displaystyle\frac{1}{3}\cdot\frac{1}{2\cdot\frac{17}{6}}\log\left|\frac{x + \frac{13}{6} - \frac{17}{6}}{x + \frac{13}{6} + \frac{17}{6}}\right| = \frac{1}{17}\log\left|\frac{3x - 2}{x + 5}\right| + C
4. (iii) Complete the square under the root.(iii) वर्गमूल के नीचे पूर्ण वर्ग बनाइए।(iii) Square root ke neeche perfect square banaiye.
   5x^2 - 2x = 5\left[\left(x - \dfrac{1}{5}\right)^2 - \left(\dfrac{1}{5}\right)^2\right] \Rightarrow \displaystyle\frac{1}{\sqrt{5}}\int \frac{dx}{\sqrt{\left(x - \frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^2}}
5. (iii) Formula (4).(iii) सूत्र (4)।(iii) Formula (4).
   = \dfrac{1}{\sqrt{5}}\log\left|x - \dfrac{1}{5} + \sqrt{x^2 - \dfrac{2x}{5}}\right| + C

✔ (i) \dfrac{1}{2}\tan^{-1}\dfrac{x - 3}{2} + C (ii) \dfrac{1}{17}\log\left|\dfrac{3x - 2}{x + 5}\right| + C (iii) \dfrac{1}{\sqrt{5}}\log\left|x - \dfrac{1}{5} + \sqrt{x^2 - \dfrac{2x}{5}}\right| + C(i) \dfrac{1}{2}\tan^{-1}\dfrac{x - 3}{2} + C (ii) \dfrac{1}{17}\log\left|\dfrac{3x - 2}{x + 5}\right| + C (iii) \dfrac{1}{\sqrt{5}}\log\left|x - \dfrac{1}{5} + \sqrt{x^2 - \dfrac{2x}{5}}\right| + C(i) \dfrac{1}{2}\tan^{-1}\dfrac{x - 3}{2} + C (ii) \dfrac{1}{17}\log\left|\dfrac{3x - 2}{x + 5}\right| + C (iii) \dfrac{1}{\sqrt{5}}\log\left|x - \dfrac{1}{5} + \sqrt{x^2 - \dfrac{2x}{5}}\right| + C

1. (i) 4x + 6 is the derivative of the denominator.(i) 4x + 6 हर का अवकलज है।(i) 4x + 6 denominator ka derivative hai.
   x + 2 = A(4x + 6) + B \Rightarrow 4A = 1,\ 6A + B = 2 \Rightarrow A = \dfrac{1}{4},\ B = \dfrac{1}{2}
2. (i) First part is f′/f; in the second, take 2 out and complete the square.(i) पहला भाग f′/f है; दूसरे में 2 बाहर निकालकर पूर्ण वर्ग बनाइए।(i) Pehla hissa f′/f hai; doosre mein 2 bahar nikaalkar perfect square banaiye.
   \displaystyle\frac{1}{4}\int \frac{4x + 6}{2x^2 + 6x + 5}\,dx + \frac{1}{2}\int \frac{dx}{2x^2 + 6x + 5} = \frac{1}{4}\log|2x^2 + 6x + 5| + \frac{1}{4}\int \frac{dx}{\left(x + \frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2}
3. (i) Formula (3) with a = 1/2.(i) a = 1/2 लेकर सूत्र (3)।(i) a = 1/2 lekar formula (3).
   = \dfrac{1}{4}\log|2x^2 + 6x + 5| + \dfrac{1}{4}\cdot\dfrac{1}{1/2}\tan^{-1}\dfrac{x + 3/2}{1/2} = \dfrac{1}{4}\log|2x^2 + 6x + 5| + \dfrac{1}{2}\tan^{-1}(2x + 3) + C
4. (ii) −4 − 2x is the derivative of 5 − 4x − x².(ii) −4 − 2x, 5 − 4x − x² का अवकलज है।(ii) −4 − 2x, 5 − 4x − x² ka derivative hai.
   x + 3 = A(-4 - 2x) + B \Rightarrow A = -\dfrac{1}{2},\ B = 1
5. (ii) First part: put t = 5 − 4x − x²; second: 5 − 4x − x² = 9 − (x + 2)², formula (5).(ii) पहला भाग: t = 5 − 4x − x² रखिए; दूसरा: 5 − 4x − x² = 9 − (x + 2)², सूत्र (5)।(ii) Pehla hissa: t = 5 − 4x − x² rakhiye; doosra: 5 − 4x − x² = 9 − (x + 2)², formula (5).
   \displaystyle -\frac{1}{2}\int \frac{-4 - 2x}{\sqrt{5 - 4x - x^2}}\,dx + \int \frac{dx}{\sqrt{9 - (x + 2)^2}} = -\sqrt{5 - 4x - x^2} + \sin^{-1}\frac{x + 2}{3} + C

✔ (i) \dfrac{1}{4}\log|2x^2 + 6x + 5| + \dfrac{1}{2}\tan^{-1}(2x + 3) + C (ii) -\sqrt{5 - 4x - x^2} + \sin^{-1}\dfrac{x + 2}{3} + C(i) \dfrac{1}{4}\log|2x^2 + 6x + 5| + \dfrac{1}{2}\tan^{-1}(2x + 3) + C (ii) -\sqrt{5 - 4x - x^2} + \sin^{-1}\dfrac{x + 2}{3} + C(i) \dfrac{1}{4}\log|2x^2 + 6x + 5| + \dfrac{1}{2}\tan^{-1}(2x + 3) + C (ii) -\sqrt{5 - 4x - x^2} + \sin^{-1}\dfrac{x + 2}{3} + C

1. Proper fraction, two different linear factors.उचित भिन्न, दो भिन्न रैखिक गुणनखंड।Proper fraction, do alag linear factors.
   \dfrac{1}{(x + 1)(x + 2)} = \dfrac{A}{x + 1} + \dfrac{B}{x + 2} \Rightarrow 1 = A(x + 2) + B(x + 1)
2. Compare the coefficients of x and the constants.x के गुणांक और अचर पदों की तुलना कीजिए।x ke coefficients aur constants compare kijiye.
   A + B = 0,\ 2A + B = 1 \Rightarrow A = 1,\ B = -1
3. Each piece is a log.हर भाग एक log है।Har piece ek log hai.
   \displaystyle\int \frac{dx}{x + 1} - \int \frac{dx}{x + 2} = \log|x + 1| - \log|x + 2| + C = \log\left|\frac{x + 1}{x + 2}\right| + C

✔ \log\left|\dfrac{x + 1}{x + 2}\right| + C\log\left|\dfrac{x + 1}{x + 2}\right| + C\log\left|\dfrac{x + 1}{x + 2}\right| + C

1. Improper (equal degrees): divide first.अनुचित (घात बराबर): पहले भाग दीजिए।Improper (degrees barabar): pehle divide kijiye.
   \dfrac{x^2 + 1}{x^2 - 5x + 6} = 1 + \dfrac{5x - 5}{(x - 2)(x - 3)}
2. Put x = 2 and x = 3 (or compare coefficients).x = 2 और x = 3 रखिए (या गुणांकों की तुलना कीजिए)।x = 2 aur x = 3 rakhiye (ya coefficients compare kijiye).
   5x - 5 = A(x - 3) + B(x - 2) \Rightarrow A = -5,\ B = 10
3. Integrate term by term.पद-दर-पद समाकलन।Term-by-term integration.
   \displaystyle\int dx - 5\int \frac{dx}{x - 2} + 10\int \frac{dx}{x - 3} = x - 5\log|x - 2| + 10\log|x - 3| + C

✔ x - 5\log|x - 2| + 10\log|x - 3| + Cx - 5\log|x - 2| + 10\log|x - 3| + Cx - 5\log|x - 2| + 10\log|x - 3| + C

1. Table 7.2 form 4 (a repeated factor).सारणी 7.2 का रूप 4 (दोहराया गया गुणनखंड)।Table 7.2 ka form 4 (repeated factor).
   3x - 2 = A(x + 1)(x + 3) + B(x + 3) + C(x + 1)^2
2. Compare coefficients of x², x and constants, and solve.x², x और अचर पदों के गुणांकों की तुलना करके हल कीजिए।x², x aur constants ke coefficients compare karke solve kijiye.
   A + C = 0,\ 4A + B + 2C = 3,\ 3A + 3B + C = -2 \Rightarrow A = \dfrac{11}{4},\ B = -\dfrac{5}{2},\ C = -\dfrac{11}{4}
3. ∫(x + 1)⁻² dx = −(x + 1)⁻¹, so the middle term turns positive.∫(x + 1)⁻² dx = −(x + 1)⁻¹, इसलिए बीच वाला पद धनात्मक हो जाता है।∫(x + 1)⁻² dx = −(x + 1)⁻¹, isliye beech wala term positive ho jaata hai.
   \displaystyle\frac{11}{4}\int \frac{dx}{x + 1} - \frac{5}{2}\int \frac{dx}{(x + 1)^2} - \frac{11}{4}\int \frac{dx}{x + 3} = \frac{11}{4}\log|x + 1| + \frac{5}{2(x + 1)} - \frac{11}{4}\log|x + 3| + C

✔ \dfrac{11}{4}\log\left|\dfrac{x + 1}{x + 3}\right| + \dfrac{5}{2(x + 1)} + C\dfrac{11}{4}\log\left|\dfrac{x + 1}{x + 3}\right| + \dfrac{5}{2(x + 1)} + C\dfrac{11}{4}\log\left|\dfrac{x + 1}{x + 3}\right| + \dfrac{5}{2(x + 1)} + C

1. Only x² appears, so treat x² as y just for splitting (y = −1 gives A, y = −4 gives B).केवल x² आता है, इसलिए तोड़ने के लिए x² को y मानिए (y = −1 से A, y = −4 से B)।Sirf x² aata hai, isliye todne ke liye x² ko y maaniye (y = −1 se A, y = −4 se B).
   x^2 = y:\ \dfrac{y}{(y + 1)(y + 4)} = \dfrac{A}{y + 1} + \dfrac{B}{y + 4} \Rightarrow A = -\dfrac{1}{3},\ B = \dfrac{4}{3}
2. Formula (3) twice.सूत्र (3) दो बार।Formula (3) do baar.
   \displaystyle -\frac{1}{3}\int \frac{dx}{x^2 + 1} + \frac{4}{3}\int \frac{dx}{x^2 + 2^2} = -\frac{1}{3}\tan^{-1}x + \frac{4}{3}\cdot\frac{1}{2}\tan^{-1}\frac{x}{2} + C

✔ -\dfrac{1}{3}\tan^{-1}x + \dfrac{2}{3}\tan^{-1}\dfrac{x}{2} + C-\dfrac{1}{3}\tan^{-1}x + \dfrac{2}{3}\tan^{-1}\dfrac{x}{2} + C-\dfrac{1}{3}\tan^{-1}x + \dfrac{2}{3}\tan^{-1}\dfrac{x}{2} + C

1. Substitution first; the denominator becomes y² − 4y + 4.पहले प्रतिस्थापन; हर बन जाता है y² − 4y + 4।Pehle substitution; denominator ban jaata hai y² − 4y + 4.
   y = \sin\phi,\ dy = \cos\phi\,d\phi:\ \displaystyle\int \frac{3y - 2}{5 - (1 - y^2) - 4y}\,dy = \int \frac{3y - 2}{(y - 2)^2}\,dy
2. Now partial fractions (form 2).अब आंशिक भिन्न (रूप 2)।Ab partial fractions (form 2).
   3y - 2 = A(y - 2) + B \Rightarrow A = 3,\ B = 4
3. Integrate both pieces.दोनों भागों का समाकलन।Dono pieces ka integration.
   \displaystyle 3\int \frac{dy}{y - 2} + 4\int \frac{dy}{(y - 2)^2} = 3\log|y - 2| - \frac{4}{y - 2} + C
4. Put y = sin φ back; 2 − sin φ is always positive, so no modulus is needed.y = sin φ वापस रखिए; 2 − sin φ सदा धनात्मक है, इसलिए मापांक की ज़रूरत नहीं।y = sin φ wapas rakhiye; 2 − sin φ hamesha positive hai, isliye modulus ki zaroorat nahi.
   = 3\log(2 - \sin\phi) + \dfrac{4}{2 - \sin\phi} + C

✔ 3\log(2 - \sin\phi) + \dfrac{4}{2 - \sin\phi} + C3\log(2 - \sin\phi) + \dfrac{4}{2 - \sin\phi} + C3\log(2 - \sin\phi) + \dfrac{4}{2 - \sin\phi} + C

1. Form 5: x² + 1 does not factorise over real numbers.रूप 5: x² + 1 के वास्तविक गुणनखंड नहीं होते।Form 5: x² + 1 ke real factors nahi hote.
   x^2 + x + 1 = A(x^2 + 1) + (Bx + C)(x + 2)
2. Compare coefficients.गुणांकों की तुलना।Coefficients compare kijiye.
   A + B = 1,\ 2B + C = 1,\ A + 2C = 1 \Rightarrow A = \dfrac{3}{5},\ B = \dfrac{2}{5},\ C = \dfrac{1}{5}
3. Split (2x + 1)/5 into an f′/f part and a tan⁻¹ part.(2x + 1)/5 को f′/f वाले भाग और tan⁻¹ वाले भाग में तोड़िए।(2x + 1)/5 ko f′/f wale hisse aur tan⁻¹ wale hisse mein todiye.
   \displaystyle\frac{3}{5}\int \frac{dx}{x + 2} + \frac{1}{5}\int \frac{2x}{x^2 + 1}\,dx + \frac{1}{5}\int \frac{dx}{x^2 + 1} = \frac{3}{5}\log|x + 2| + \frac{1}{5}\log(x^2 + 1) + \frac{1}{5}\tan^{-1}x + C

✔ \dfrac{3}{5}\log|x + 2| + \dfrac{1}{5}\log(x^2 + 1) + \dfrac{1}{5}\tan^{-1}x + C\dfrac{3}{5}\log|x + 2| + \dfrac{1}{5}\log(x^2 + 1) + \dfrac{1}{5}\tan^{-1}x + C\dfrac{3}{5}\log|x + 2| + \dfrac{1}{5}\log(x^2 + 1) + \dfrac{1}{5}\tan^{-1}x + C

1. x becomes 1 on differentiation, so it is the first function.x अवकलन से 1 हो जाता है, इसलिए यह प्रथम फलन है।x differentiate karne se 1 ho jaata hai, isliye ye first function hai.
   f_1 = x,\ f_2 = \cos x:\ \displaystyle\int x\cos x\,dx = x\int \cos x\,dx - \int\left[\frac{d}{dx}(x)\int \cos x\,dx\right]dx
2. The leftover integral is standard.बचा हुआ समाकल मानक है।Bacha hua integral standard hai.
   = x\sin x - \displaystyle\int \sin x\,dx = x\sin x + \cos x + C

✔ x\sin x + \cos x + Cx\sin x + \cos x + Cx\sin x + \cos x + C

1. We cannot guess a function whose derivative is log x, so take 1 as the second function.log x का अवकलज जिस फलन का हो, उसका अनुमान नहीं लगता, इसलिए 1 को द्वितीय फलन लीजिए।Kis function ka derivative log x hai, ye guess nahi hota, isliye 1 ko second function lijiye.
   \displaystyle\int \log x\cdot 1\,dx = \log x\int 1\,dx - \int\left[\frac{d}{dx}(\log x)\int 1\,dx\right]dx
2. (1/x) · x = 1.(1/x) · x = 1।(1/x) · x = 1.
   = (\log x)\cdot x - \displaystyle\int \frac{1}{x}\cdot x\,dx = x\log x - x + C

✔ x\log x - x + Cx\log x - x + Cx\log x - x + C

1. x first, eˣ second (its integral is itself).x प्रथम, eˣ द्वितीय (इसका समाकल यही है)।x first, eˣ second (iska integral khud yahi hai).
   \displaystyle\int x e^x\,dx = x e^x - \int 1\cdot e^x\,dx = x e^x - e^x + C

✔ x e^x - e^x + Cx e^x - e^x + Cx e^x - e^x + C

1. First find the integral of the second function x/√(1 − x²); first function = sin⁻¹x (inverse trigonometric).पहले द्वितीय फलन x/√(1 − x²) का समाकल निकालिए; प्रथम फलन = sin⁻¹x (प्रतिलोम त्रिकोणमितीय)।Pehle second function x/√(1 − x²) ka integral nikaaliye; first function = sin⁻¹x (inverse trigonometric).
   \displaystyle\int \frac{x\,dx}{\sqrt{1 - x^2}}:\ t = 1 - x^2,\ dt = -2x\,dx \Rightarrow -\frac{1}{2}\int \frac{dt}{\sqrt{t}} = -\sqrt{t} = -\sqrt{1 - x^2}
2. By parts.खंडशः समाकलन।By parts.
   \displaystyle\int \frac{x\sin^{-1}x}{\sqrt{1 - x^2}}\,dx = (\sin^{-1}x)(-\sqrt{1 - x^2}) - \int \frac{1}{\sqrt{1 - x^2}}(-\sqrt{1 - x^2})\,dx
3. The two square roots cancel, leaving ∫1 dx.दोनों वर्गमूल कट जाते हैं, ∫1 dx बचता है।Dono square roots cut jaate hain, ∫1 dx bachta hai.
   = -\sqrt{1 - x^2}\,\sin^{-1}x + x + C

✔ -\sqrt{1 - x^2}\,\sin^{-1}x + x + C-\sqrt{1 - x^2}\,\sin^{-1}x + x + C-\sqrt{1 - x^2}\,\sin^{-1}x + x + C

1. eˣ first, sin x second.eˣ प्रथम, sin x द्वितीय।eˣ first, sin x second.
   I = \displaystyle\int e^x\sin x\,dx = e^x(-\cos x) + \int e^x\cos x\,dx
2. By parts once more, with the same choice.एक बार फिर खंडशः, वही चुनाव रखकर।Ek baar phir by parts, wahi chunav rakhkar.
   \displaystyle\int e^x\cos x\,dx = e^x\sin x - \int e^x\sin x\,dx = e^x\sin x - I
3. The original integral comes back — move it to the left side.मूल समाकल वापस आ गया — उसे बाईं ओर ले जाइए।Original integral wapas aa gaya — use left side le jaaiye.
   I = -e^x\cos x + e^x\sin x - I \Rightarrow 2I = e^x(\sin x - \cos x)

✔ \dfrac{e^x}{2}(\sin x - \cos x) + C\dfrac{e^x}{2}(\sin x - \cos x) + C\dfrac{e^x}{2}(\sin x - \cos x) + C

1. (i) The bracket is exactly f + f′.(i) कोष्ठक ठीक f + f′ है।(i) Bracket bilkul f + f′ hai.
   f(x) = \tan^{-1}x,\ f'(x) = \dfrac{1}{1 + x^2} \Rightarrow \displaystyle\int = e^x\tan^{-1}x + C
2. (ii) Rewrite x² + 1 as (x² − 1) + 2.(ii) x² + 1 को (x² − 1) + 2 लिखिए।(ii) x² + 1 ko (x² − 1) + 2 likhiye.
   \dfrac{x^2 + 1}{(x + 1)^2} = \dfrac{x^2 - 1 + 2}{(x + 1)^2} = \dfrac{x - 1}{x + 1} + \dfrac{2}{(x + 1)^2}
3. (ii) Now it is eˣ(f + f′).(ii) अब यह eˣ(f + f′) है।(ii) Ab ye eˣ(f + f′) hai.
   f(x) = \dfrac{x - 1}{x + 1},\ f'(x) = \dfrac{2}{(x + 1)^2} \Rightarrow \displaystyle\int = \frac{x - 1}{x + 1}e^x + C

✔ (i) e^x\tan^{-1}x + C (ii) \dfrac{x - 1}{x + 1}e^x + C(i) e^x\tan^{-1}x + C (ii) \dfrac{x - 1}{x + 1}e^x + C(i) e^x\tan^{-1}x + C (ii) \dfrac{x - 1}{x + 1}e^x + C

1. Complete the square.पूर्ण वर्ग बनाइए।Perfect square banaiye.
   x^2 + 2x + 5 = (x + 1)^2 + 2^2;\ x + 1 = y,\ dx = dy:\ \displaystyle\int \sqrt{y^2 + 2^2}\,dy
2. Result (ii) with a = 2.a = 2 लेकर परिणाम (ii)।a = 2 lekar result (ii).
   = \dfrac{y}{2}\sqrt{y^2 + 4} + \dfrac{4}{2}\log\left|y + \sqrt{y^2 + 4}\right| + C

✔ \dfrac{1}{2}(x + 1)\sqrt{x^2 + 2x + 5} + 2\log\left|x + 1 + \sqrt{x^2 + 2x + 5}\right| + C\dfrac{1}{2}(x + 1)\sqrt{x^2 + 2x + 5} + 2\log\left|x + 1 + \sqrt{x^2 + 2x + 5}\right| + C\dfrac{1}{2}(x + 1)\sqrt{x^2 + 2x + 5} + 2\log\left|x + 1 + \sqrt{x^2 + 2x + 5}\right| + C

1. Complete the square.पूर्ण वर्ग बनाइए।Perfect square banaiye.
   3 - 2x - x^2 = 4 - (x + 1)^2;\ x + 1 = y:\ \displaystyle\int \sqrt{2^2 - y^2}\,dy
2. Result (iii) with a = 2.a = 2 लेकर परिणाम (iii)।a = 2 lekar result (iii).
   = \dfrac{y}{2}\sqrt{4 - y^2} + \dfrac{4}{2}\sin^{-1}\dfrac{y}{2} + C

✔ \dfrac{1}{2}(x + 1)\sqrt{3 - 2x - x^2} + 2\sin^{-1}\dfrac{x + 1}{2} + C\dfrac{1}{2}(x + 1)\sqrt{3 - 2x - x^2} + 2\sin^{-1}\dfrac{x + 1}{2} + C\dfrac{1}{2}(x + 1)\sqrt{3 - 2x - x^2} + 2\sin^{-1}\dfrac{x + 1}{2} + C

1. (i) F(x) = x³/3.(i) F(x) = x³/3।(i) F(x) = x³/3.
   \left[\dfrac{x^3}{3}\right]_2^3 = \dfrac{27}{3} - \dfrac{8}{3} = \dfrac{19}{3}
2. (ii) Find the anti derivative by substitution first.(ii) पहले प्रतिस्थापन से प्रतिअवकलज निकालिए।(ii) Pehle substitution se anti derivative nikaaliye.
   t = 30 - x^{3/2},\ dt = -\tfrac{3}{2}\sqrt{x}\,dx:\ F(x) = \dfrac{2}{3}\cdot\dfrac{1}{30 - x^{3/2}}
3. (ii) 9^3/2 = 27, 4^3/2 = 8.(ii) 9^3/2 = 27, 4^3/2 = 8।(ii) 9^3/2 = 27, 4^3/2 = 8.
   F(9) - F(4) = \dfrac{2}{3}\left[\dfrac{1}{30 - 27} - \dfrac{1}{30 - 8}\right] = \dfrac{2}{3}\left[\dfrac{1}{3} - \dfrac{1}{22}\right] = \dfrac{19}{99}
4. (iii) Partial fractions.(iii) आंशिक भिन्न।(iii) Partial fractions.
   \dfrac{x}{(x + 1)(x + 2)} = \dfrac{-1}{x + 1} + \dfrac{2}{x + 2} \Rightarrow F(x) = -\log|x + 1| + 2\log|x + 2|
5. (iii) Combine the logs: 2 × 16 / 27.(iii) log को मिलाइए: 2 × 16 / 27।(iii) Logs ko milaiye: 2 × 16 / 27.
   F(2) - F(1) = [-\log 3 + 2\log 4] - [-\log 2 + 2\log 3] = -3\log 3 + \log 2 + 2\log 4 = \log\dfrac{32}{27}
6. (iv) sin(π/2) = 1, sin 0 = 0.(iv) sin(π/2) = 1, sin 0 = 0।(iv) sin(π/2) = 1, sin 0 = 0.
   u = \sin 2t,\ du = 2\cos 2t\,dt:\ F(t) = \dfrac{\sin^4 2t}{8};\ F\left(\dfrac{\pi}{4}\right) - F(0) = \dfrac{1}{8} - 0

✔ (i) \dfrac{19}{3} (ii) \dfrac{19}{99} (iii) \log\dfrac{32}{27} (iv) \dfrac{1}{8}(i) \dfrac{19}{3} (ii) \dfrac{19}{99} (iii) \log\dfrac{32}{27} (iv) \dfrac{1}{8}(i) \dfrac{19}{3} (ii) \dfrac{19}{99} (iii) \log\dfrac{32}{27} (iv) \dfrac{1}{8}

1. Substitute and change the limits.प्रतिस्थापन कीजिए और सीमाएँ बदलिए।Substitute kijiye aur limits badliye.
   t = x^5 + 1,\ dt = 5x^4\,dx;\ x = -1 \Rightarrow t = 0,\ x = 1 \Rightarrow t = 2
2. 2^3/2 = 2√2.2^3/2 = 2√2।2^3/2 = 2√2.
   \displaystyle\int_0^2 \sqrt{t}\,dt = \left[\frac{2}{3}t^{3/2}\right]_0^2 = \frac{2}{3}\cdot 2\sqrt{2} = \frac{4\sqrt{2}}{3}

✔ \dfrac{4\sqrt{2}}{3}\dfrac{4\sqrt{2}}{3}\dfrac{4\sqrt{2}}{3}

1. The derivative of tan⁻¹x is present.tan⁻¹x का अवकलज मौजूद है।tan⁻¹x ka derivative maujood hai.
   t = \tan^{-1}x,\ dt = \dfrac{dx}{1 + x^2};\ x = 0 \Rightarrow t = 0,\ x = 1 \Rightarrow t = \dfrac{\pi}{4}
2. Power rule.घात का नियम।Power rule.
   \displaystyle\int_0^{\pi/4} t\,dt = \left[\frac{t^2}{2}\right]_0^{\pi/4} = \frac{1}{2}\cdot\frac{\pi^2}{16} = \frac{\pi^2}{32}

✔ \dfrac{\pi^2}{32}\dfrac{\pi^2}{32}\dfrac{\pi^2}{32}

1. Find where the expression inside | | changes sign.पता कीजिए कि | | के अंदर का व्यंजक कहाँ चिह्न बदलता है।Pata kijiye ki | | ke andar ka expression kahan sign badalta hai.
   x^3 - x = x(x - 1)(x + 1):\ \ge 0 \text{ on } [-1, 0],\ \le 0 \text{ on } [0, 1],\ \ge 0 \text{ on } [1, 2]
2. P₂: split at 0 and 1; where it is negative, |x³ − x| = −(x³ − x).P₂: 0 और 1 पर तोड़िए; जहाँ ऋणात्मक है, वहाँ |x³ − x| = −(x³ − x)।P₂: 0 aur 1 par todiye; jahan negative hai, wahan |x³ − x| = −(x³ − x).
   \displaystyle\int_{-1}^{0}(x^3 - x)\,dx - \int_0^1 (x^3 - x)\,dx + \int_1^2 (x^3 - x)\,dx
3. Evaluate each piece.हर भाग का मान निकालिए।Har piece ki value nikaaliye.
   = \left[\dfrac{x^4}{4} - \dfrac{x^2}{2}\right]_{-1}^{0} - \left[\dfrac{x^4}{4} - \dfrac{x^2}{2}\right]_0^1 + \left[\dfrac{x^4}{4} - \dfrac{x^2}{2}\right]_1^2 = \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{9}{4}

✔ \dfrac{11}{4}\dfrac{11}{4}\dfrac{11}{4}

1. sin² x is even, so use P₇(i).sin² x सम फलन है, इसलिए P₇(i)।sin² x even hai, isliye P₇(i).
   \sin^2(-x) = \sin^2 x \Rightarrow \displaystyle\int_{-\pi/4}^{\pi/4} \sin^2 x\,dx = 2\int_0^{\pi/4} \sin^2 x\,dx
2. 2 sin² x = 1 − cos 2x.2 sin² x = 1 − cos 2x।2 sin² x = 1 − cos 2x.
   = \displaystyle\int_0^{\pi/4}(1 - \cos 2x)\,dx = \left[x - \frac{\sin 2x}{2}\right]_0^{\pi/4} = \frac{\pi}{4} - \frac{1}{2}

✔ \dfrac{\pi}{4} - \dfrac{1}{2}\dfrac{\pi}{4} - \dfrac{1}{2}\dfrac{\pi}{4} - \dfrac{1}{2}

1. P₄: replace x by π − x; sin(π − x) = sin x, cos²(π − x) = cos² x.P₄: x की जगह π − x; sin(π − x) = sin x, cos²(π − x) = cos² x।P₄: x ki jagah π − x; sin(π − x) = sin x, cos²(π − x) = cos² x.
   I = \displaystyle\int_0^{\pi} \frac{(\pi - x)\sin(\pi - x)}{1 + \cos^2(\pi - x)}\,dx = \int_0^{\pi} \frac{(\pi - x)\sin x}{1 + \cos^2 x}\,dx
2. Add the two forms of I: the x terms cancel.I के दोनों रूप जोड़िए: x वाले पद कट जाते हैं।I ke dono roop jodiye: x wale terms cancel ho jaate hain.
   2I = \pi\displaystyle\int_0^{\pi} \frac{\sin x}{1 + \cos^2 x}\,dx
3. x = 0 → t = 1, x = π → t = −1; flipping the limits removes the minus.x = 0 → t = 1, x = π → t = −1; सीमाएँ पलटने से ऋण चिह्न हट जाता है।x = 0 → t = 1, x = π → t = −1; limits palatne se minus hat jaata hai.
   \cos x = t,\ -\sin x\,dx = dt:\ 2I = \pi\displaystyle\int_{-1}^{1} \frac{dt}{1 + t^2} = \pi\left[\tan^{-1}t\right]_{-1}^{1} = \pi\cdot\frac{\pi}{2}

✔ I = \dfrac{\pi^2}{4}I = \dfrac{\pi^2}{4}I = \dfrac{\pi^2}{4}

1. f is odd.f विषम फलन है।f odd function hai.
   f(-x) = \sin^5(-x)\cos^4(-x) = -\sin^5 x\cos^4 x = -f(x)
2. P₇(ii).P₇(ii)।P₇(ii).
   \displaystyle\int_{-1}^{1} f(x)\,dx = 0

✔ 000

1. P₄ with x → π/2 − x: sin and cos swap.P₄ में x → π/2 − x: sin और cos आपस में बदल जाते हैं।P₄ mein x → π/2 − x: sin aur cos aapas mein badal jaate hain.
   I = \displaystyle\int_0^{\pi/2} \frac{\cos^4 x}{\cos^4 x + \sin^4 x}\,dx
2. Add the two forms.दोनों रूप जोड़िए।Dono roop jodiye.
   2I = \displaystyle\int_0^{\pi/2} \frac{\sin^4 x + \cos^4 x}{\sin^4 x + \cos^4 x}\,dx = \int_0^{\pi/2} dx = \frac{\pi}{2}

✔ I = \dfrac{\pi}{4}I = \dfrac{\pi}{4}I = \dfrac{\pi}{4}

1. Write tan x = sin x / cos x.tan x = sin x / cos x लिखिए।tan x = sin x / cos x likhiye.
   I = \displaystyle\int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}\,dx
2. P₃: a + b − x = π/2 − x, which swaps sin and cos.P₃: a + b − x = π/2 − x, जिससे sin और cos आपस में बदल जाते हैं।P₃: a + b − x = π/2 − x, jisse sin aur cos aapas mein badal jaate hain.
   I = \displaystyle\int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx
3. Add.जोड़िए।Jodiye.
   2I = \displaystyle\int_{\pi/6}^{\pi/3} dx = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}

✔ I = \dfrac{\pi}{12}I = \dfrac{\pi}{12}I = \dfrac{\pi}{12}

1. P₄, then add the two forms.P₄, फिर दोनों रूप जोड़िए।P₄, phir dono roop jodiye.
   I = \displaystyle\int_0^{\pi/2} \log\cos x\,dx \Rightarrow 2I = \int_0^{\pi/2} \log(\sin x\cos x)\,dx
2. sin x cos x = (sin 2x)/2.sin x cos x = (sin 2x)/2।sin x cos x = (sin 2x)/2.
   2I = \displaystyle\int_0^{\pi/2} \log\frac{\sin 2x}{2}\,dx = \int_0^{\pi/2} \log\sin 2x\,dx - \frac{\pi}{2}\log 2
3. P₆: sin(π − t) = sin t, so the integral over [0, π] is twice the one over [0, π/2].P₆: sin(π − t) = sin t, इसलिए [0, π] पर समाकल, [0, π/2] वाले का दुगुना है।P₆: sin(π − t) = sin t, isliye [0, π] par integral, [0, π/2] wale ka double hai.
   2x = t:\ \displaystyle\int_0^{\pi/2} \log\sin 2x\,dx = \frac{1}{2}\int_0^{\pi} \log\sin t\,dt = \frac{1}{2}\cdot 2\int_0^{\pi/2} \log\sin t\,dt = I
4. Substitute back.वापस रखिए।Wapas rakhiye.
   2I = I - \dfrac{\pi}{2}\log 2

✔ I = -\dfrac{\pi}{2}\log 2I = -\dfrac{\pi}{2}\log 2I = -\dfrac{\pi}{2}\log 2

1. cos 6x is (1/6) of the derivative of 1 + sin 6x.cos 6x, 1 + sin 6x के अवकलज का 1/6 है।cos 6x, 1 + sin 6x ke derivative ka 1/6 hai.
   t = 1 + \sin 6x,\ dt = 6\cos 6x\,dx:\ \displaystyle\frac{1}{6}\int t^{1/2}\,dt = \frac{1}{6}\cdot\frac{2}{3}t^{3/2} + C

✔ \dfrac{1}{9}(1 + \sin 6x)^{3/2} + C\dfrac{1}{9}(1 + \sin 6x)^{3/2} + C\dfrac{1}{9}(1 + \sin 6x)^{3/2} + C

1. Take x⁴ out of the fourth root.चतुर्थ मूल से x⁴ बाहर निकालिए।Fourth root se x⁴ bahar nikaaliye.
   \dfrac{(x^4 - x)^{1/4}}{x^5} = \dfrac{x\left(1 - \frac{1}{x^3}\right)^{1/4}}{x^5} = \dfrac{1}{x^4}\left(1 - \dfrac{1}{x^3}\right)^{1/4}
2. The derivative of 1 − x⁻³ is 3x⁻⁴.1 − x⁻³ का अवकलज 3x⁻⁴ है।1 − x⁻³ ka derivative 3x⁻⁴ hai.
   1 - \dfrac{1}{x^3} = t,\ \dfrac{3}{x^4}\,dx = dt:\ \displaystyle\frac{1}{3}\int t^{1/4}\,dt = \frac{1}{3}\cdot\frac{4}{5}t^{5/4} + C

✔ \dfrac{4}{15}\left(1 - \dfrac{1}{x^3}\right)^{5/4} + C\dfrac{4}{15}\left(1 - \dfrac{1}{x^3}\right)^{5/4} + C\dfrac{4}{15}\left(1 - \dfrac{1}{x^3}\right)^{5/4} + C

1. Improper: divide x⁴ by x³ − x² + x − 1.अनुचित: x⁴ को x³ − x² + x − 1 से भाग दीजिए।Improper: x⁴ ko x³ − x² + x − 1 se divide kijiye.
   \dfrac{x^4}{(x - 1)(x^2 + 1)} = (x + 1) + \dfrac{1}{x^3 - x^2 + x - 1}
2. Partial fractions (form 5).आंशिक भिन्न (रूप 5)।Partial fractions (form 5).
   \dfrac{1}{(x - 1)(x^2 + 1)} = \dfrac{A}{x - 1} + \dfrac{Bx + C}{x^2 + 1} \Rightarrow A = \dfrac{1}{2},\ B = C = -\dfrac{1}{2}
3. Split −(x + 1)/(2(x² + 1)) into an f′/f part and a tan⁻¹ part.−(x + 1)/(2(x² + 1)) को f′/f भाग और tan⁻¹ भाग में तोड़िए।−(x + 1)/(2(x² + 1)) ko f′/f hisse aur tan⁻¹ hisse mein todiye.
   \displaystyle\int (x + 1)\,dx + \frac{1}{2}\int \frac{dx}{x - 1} - \frac{1}{4}\int \frac{2x\,dx}{x^2 + 1} - \frac{1}{2}\int \frac{dx}{x^2 + 1}

✔ \dfrac{x^2}{2} + x + \dfrac{1}{2}\log|x - 1| - \dfrac{1}{4}\log(x^2 + 1) - \dfrac{1}{2}\tan^{-1}x + C\dfrac{x^2}{2} + x + \dfrac{1}{2}\log|x - 1| - \dfrac{1}{4}\log(x^2 + 1) - \dfrac{1}{2}\tan^{-1}x + C\dfrac{x^2}{2} + x + \dfrac{1}{2}\log|x - 1| - \dfrac{1}{4}\log(x^2 + 1) - \dfrac{1}{2}\tan^{-1}x + C

1. By parts with 1 as the second function.1 को द्वितीय फलन लेकर खंडशः।1 ko second function lekar by parts.
   \displaystyle\int \log(\log x)\cdot 1\,dx = x\log(\log x) - \int x\cdot\frac{1}{x\log x}\,dx = x\log(\log x) - \int \frac{dx}{\log x}
2. By parts again: the derivative of 1/log x is −1/(x (log x)²).फिर से खंडशः: 1/log x का अवकलज −1/(x (log x)²) है।Phir se by parts: 1/log x ka derivative −1/(x (log x)²) hai.
   \displaystyle\int \frac{dx}{\log x} = \frac{x}{\log x} + \int \frac{dx}{(\log x)^2}
3. The two leftover integrals cancel.बचे हुए दोनों समाकल कट जाते हैं।Bache hue dono integrals cancel ho jaate hain.
   I = x\log(\log x) - \dfrac{x}{\log x} - \displaystyle\int \frac{dx}{(\log x)^2} + \int \frac{dx}{(\log x)^2}

✔ x\log(\log x) - \dfrac{x}{\log x} + Cx\log(\log x) - \dfrac{x}{\log x} + Cx\log(\log x) - \dfrac{x}{\log x} + C

1. Write everything in √tan x.सब कुछ √tan x में लिखिए।Sab kuch √tan x mein likhiye.
   \sqrt{\tan x} = t,\ \sec^2 x\,dx = 2t\,dt \Rightarrow dx = \dfrac{2t\,dt}{1 + t^4}
2. √cot x + √tan x = √tan x (1 + cot x); divide top and bottom by t².√cot x + √tan x = √tan x (1 + cot x); अंश और हर को t² से भाग दीजिए।√cot x + √tan x = √tan x (1 + cot x); numerator aur denominator ko t² se divide kijiye.
   I = \displaystyle\int t\left(1 + \frac{1}{t^2}\right)\frac{2t}{1 + t^4}\,dt = 2\int \frac{t^2 + 1}{t^4 + 1}\,dt = 2\int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}}\,dt
3. t² + 1/t² = (t − 1/t)² + 2.t² + 1/t² = (t − 1/t)² + 2।t² + 1/t² = (t − 1/t)² + 2.
   t - \dfrac{1}{t} = y,\ \left(1 + \dfrac{1}{t^2}\right)dt = dy:\ \displaystyle 2\int \frac{dy}{y^2 + (\sqrt{2})^2} = \sqrt{2}\tan^{-1}\frac{y}{\sqrt{2}} + C

✔ \sqrt{2}\tan^{-1}\left(\dfrac{\tan x - 1}{\sqrt{2\tan x}}\right) + C\sqrt{2}\tan^{-1}\left(\dfrac{\tan x - 1}{\sqrt{2\tan x}}\right) + C\sqrt{2}\tan^{-1}\left(\dfrac{\tan x - 1}{\sqrt{2\tan x}}\right) + C

1. cos⁴ 2x = t², then formula (5) with a = 3.cos⁴ 2x = t², फिर a = 3 लेकर सूत्र (5)।cos⁴ 2x = t², phir a = 3 lekar formula (5).
   \cos^2(2x) = t,\ -4\sin 2x\cos 2x\,dx = dt:\ \displaystyle -\frac{1}{4}\int \frac{dt}{\sqrt{9 - t^2}} = -\frac{1}{4}\sin^{-1}\frac{t}{3} + C

✔ -\dfrac{1}{4}\sin^{-1}\left[\dfrac{1}{3}\cos^2 2x\right] + C-\dfrac{1}{4}\sin^{-1}\left[\dfrac{1}{3}\cos^2 2x\right] + C-\dfrac{1}{4}\sin^{-1}\left[\dfrac{1}{3}\cos^2 2x\right] + C

1. On [−1, 1] x and sin πx have the same sign; on [1, 3/2] sin πx ≤ 0 while x > 0.[−1, 1] पर x और sin πx का चिह्न समान है; [1, 3/2] पर sin πx ≤ 0 जबकि x > 0।[−1, 1] par x aur sin πx ka sign same hai; [1, 3/2] par sin πx ≤ 0 jabki x > 0.
   x\sin\pi x \ge 0 \text{ on } [-1, 1],\ \le 0 \text{ on } \left[1, \tfrac{3}{2}\right] \Rightarrow I = \displaystyle\int_{-1}^{1} x\sin\pi x\,dx - \int_1^{3/2} x\sin\pi x\,dx
2. By parts, x as the first function.खंडशः, x प्रथम फलन।By parts, x first function.
   \displaystyle\int x\sin\pi x\,dx = -\frac{x\cos\pi x}{\pi} + \frac{\sin\pi x}{\pi^2}
3. cos π = −1, cos(3π/2) = 0, sin(3π/2) = −1.cos π = −1, cos(3π/2) = 0, sin(3π/2) = −1।cos π = −1, cos(3π/2) = 0, sin(3π/2) = −1.
   \left[-\dfrac{x\cos\pi x}{\pi} + \dfrac{\sin\pi x}{\pi^2}\right]_{-1}^{1} = \dfrac{2}{\pi},\quad \left[-\dfrac{x\cos\pi x}{\pi} + \dfrac{\sin\pi x}{\pi^2}\right]_{1}^{3/2} = -\dfrac{1}{\pi^2} - \dfrac{1}{\pi}
4. Combine.मिलाइए।Milaiye.
   I = \dfrac{2}{\pi} - \left(-\dfrac{1}{\pi^2} - \dfrac{1}{\pi}\right)

✔ \dfrac{3}{\pi} + \dfrac{1}{\pi^2}\dfrac{3}{\pi} + \dfrac{1}{\pi^2}\dfrac{3}{\pi} + \dfrac{1}{\pi^2}

1. P₄ (x → π − x) and add; then P₆ because the integrand is unchanged by x → π − x.P₄ (x → π − x) और जोड़िए; फिर P₆, क्योंकि x → π − x से समाकल्य नहीं बदलता।P₄ (x → π − x) aur jodiye; phir P₆, kyunki x → π − x se integrand nahi badalta.
   2I = \pi\displaystyle\int_0^{\pi} \frac{dx}{a^2\cos^2 x + b^2\sin^2 x} = 2\pi\int_0^{\pi/2} \frac{dx}{a^2\cos^2 x + b^2\sin^2 x}
2. Split at π/4; divide by cos² x in the first part and by sin² x in the second.π/4 पर तोड़िए; पहले भाग में cos² x से और दूसरे में sin² x से भाग दीजिए।π/4 par todiye; pehle hisse mein cos² x se aur doosre mein sin² x se divide kijiye.
   I = \pi\left[\displaystyle\int_0^{\pi/4} \frac{\sec^2 x\,dx}{a^2 + b^2\tan^2 x} + \int_{\pi/4}^{\pi/2} \frac{\operatorname{cosec}^2 x\,dx}{a^2\cot^2 x + b^2}\right]
3. Formula (3) in each part (a, b > 0).हर भाग में सूत्र (3) (a, b > 0)।Har hisse mein formula (3) (a, b > 0).
   \tan x = t,\ \cot x = u:\ I = \pi\left[\displaystyle\int_0^1 \frac{dt}{a^2 + b^2t^2} + \int_0^1 \frac{du}{a^2u^2 + b^2}\right] = \frac{\pi}{ab}\left[\tan^{-1}\frac{b}{a} + \tan^{-1}\frac{a}{b}\right]
4. tan⁻¹y + cot⁻¹y = π/2 for y > 0.y > 0 के लिए tan⁻¹y + cot⁻¹y = π/2।y > 0 ke liye tan⁻¹y + cot⁻¹y = π/2.
   \tan^{-1}\dfrac{b}{a} + \tan^{-1}\dfrac{a}{b} = \dfrac{\pi}{2}

✔ \dfrac{\pi^2}{2ab}\dfrac{\pi^2}{2ab}\dfrac{\pi^2}{2ab}