Miscellaneous Exercise questions and solutions
Miscellaneous Exercise, Question 1
\(\displaystyle\int \dfrac{1}{x - x^3}\,dx\)
Show solution
- \[\dfrac{1}{x(1 - x)(1 + x)} = \dfrac1x + \dfrac{1/2}{1 - x} - \dfrac{1/2}{1 + x}\]
- \[\log|x| - \tfrac12\log|1 - x| - \tfrac12\log|1 + x|\]
Answer: \[\tfrac12\log\left|\dfrac{x^2}{1 - x^2}\right| + C\]
Practise this: Step 2, Board standard →
Miscellaneous Exercise, Question 2
\(\displaystyle\int \dfrac{1}{\sqrt{x + a} + \sqrt{x + b}}\,dx\)
Show solution
- Rationalise: \[\dfrac{\sqrt{x + a} - \sqrt{x + b}}{(x + a) - (x + b)} = \dfrac{\sqrt{x + a} - \sqrt{x + b}}{a - b}\]
Answer: \[\dfrac{2}{3(a - b)}\left[(x + a)^{3/2} - (x + b)^{3/2}\right] + C\]
Practise this: Step 2, Board standard →
Miscellaneous Exercise, Question 3
\(\displaystyle\int \dfrac{1}{x\sqrt{ax - x^2}}\,dx\)
Show solution
- \(x = \dfrac{a}{t}\), \(dx = -\dfrac{a}{t^2}\,dt\); \(ax - x^2 = \dfrac{a^2(t - 1)}{t^2}\).
- The integral becomes \[-\dfrac1a\int(t - 1)^{-1/2}\,dt = -\dfrac2a\sqrt{t - 1}\], with \(t - 1 = \dfrac{a - x}{x}\).
Answer: \(-\dfrac2a\sqrt{\dfrac{a - x}{x}} + C\)
Practise this: Step 3, Full marks on long answers →
Miscellaneous Exercise, Question 4
\(\displaystyle\int \dfrac{1}{x^2(x^4 + 1)^{3/4}}\,dx\)
Show solution
- For \(x > 0\): \[\begin{aligned}x^2(x^4 + 1)^{3/4} &= x^2 \cdot x^3(1 + x^{-4})^{3/4} \\ &= x^5(1 + x^{-4})^{3/4}\end{aligned}\]
- \(t = 1 + x^{-4}\), \(dt = -4x^{-5}\,dx\): \(-\tfrac14\int t^{-3/4}\,dt = -t^{1/4}\).
Answer: \[-\left(1 + \dfrac{1}{x^4}\right)^{1/4} + C\]
Practise this: Step 3, Full marks on long answers →
Miscellaneous Exercise, Question 5
\(\displaystyle\int \dfrac{1}{x^{1/2} + x^{1/3}}\,dx\)
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- \(x = t^6\), \(dx = 6t^5\,dt\): \[\int\dfrac{6t^5}{t^3 + t^2}\,dt = 6\int\dfrac{t^3}{t + 1}\,dt\]
- \[\dfrac{t^3}{t + 1} = t^2 - t + 1 - \dfrac{1}{t + 1}\]
Answer: \[2\sqrt x - 3x^{1/3} + 6x^{1/6} - 6\log(1 + x^{1/6}) + C\]
Practise this: Step 3, Full marks on long answers →
Miscellaneous Exercise, Question 6
\(\displaystyle\int \dfrac{5x}{(x + 1)(x^2 + 9)}\,dx\)
Show solution
- \[\dfrac{A}{x + 1} + \dfrac{Bx + C}{x^2 + 9}\]: \(A = -\tfrac12\); \(x^2\): \(A + B = 0\); constant: \(9A + C = 0\). So \(B = \tfrac12\), \(C = \tfrac92\).
Answer: \[-\tfrac12\log|x + 1| + \tfrac14\log(x^2 + 9) + \tfrac32\tan^{-1}\dfrac{x}{3} + C\]
Practise this: Step 2, Board standard →
Miscellaneous Exercise, Question 7
\(\displaystyle\int \dfrac{\sin x}{\sin(x - a)}\,dx\)
Show solution
- \(x - a = t\): \[\dfrac{\sin(t + a)}{\sin t} = \cos a + \sin a\cot t\]
- \(t\cos a + \sin a\log|\sin t|\); the \(-a\cos a\) goes into C.
Answer: \(x\cos a + \sin a\log|\sin(x - a)| + C\)
Practise this: Step 3, Full marks on long answers →
Miscellaneous Exercise, Question 8
\(\displaystyle\int \dfrac{e^{5\log x} - e^{4\log x}}{e^{3\log x} - e^{2\log x}}\,dx\)
Show solution
- \(e^{k\log x} = x^k\): \[\begin{aligned}\dfrac{x^5 - x^4}{x^3 - x^2} &= \dfrac{x^4(x - 1)}{x^2(x - 1)} \\ &= x^2\end{aligned}\]
Answer: \(\dfrac{x^3}{3} + C\)
Practise this: Step 1, Secure the basics →
Miscellaneous Exercise, Question 9
\(\displaystyle\int \dfrac{\cos x}{\sqrt{4 - \sin^2 x}}\,dx\)
Show solution
- \(t = \sin x\): \(\int\dfrac{dt}{\sqrt{2^2 - t^2}}\).
Answer: \[\sin^{-1}\left(\dfrac{\sin x}{2}\right) + C\]
Practise this: Step 1, Secure the basics →
Miscellaneous Exercise, Question 10
\(\displaystyle\int \dfrac{\sin^8 x - \cos^8 x}{1 - 2\sin^2 x\cos^2 x}\,dx\)
Show solution
- \[\sin^8 x - \cos^8 x = (\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)\] and \[\sin^4 x + \cos^4 x = 1 - 2\sin^2 x\cos^2 x\]
- Left: \[\begin{aligned}\sin^4 x - \cos^4 x &= \sin^2 x - \cos^2 x \\ &= -\cos 2x\end{aligned}\]
Answer: \(-\tfrac12\sin 2x + C\)
Practise this: Step 2, Board standard →
Miscellaneous Exercise, Question 11
\(\displaystyle\int \dfrac{1}{\cos(x + a)\cos(x + b)}\,dx\)
Show solution
- \[1 = \dfrac{\sin[(x + a) - (x + b)]}{\sin(a - b)}\]; expanding gives \[\dfrac{\tan(x + a) - \tan(x + b)}{\sin(a - b)}\]
- \(\int\tan(x + a)\,dx = -\log|\cos(x + a)|\).
Answer: \[\dfrac{1}{\sin(a - b)}\log\left|\dfrac{\cos(x + b)}{\cos(x + a)}\right| + C\]
Practise this: Step 4, 95+ stretch (HOTS) →
Miscellaneous Exercise, Question 12
\(\displaystyle\int \dfrac{x^3}{\sqrt{1 - x^8}}\,dx\)
Show solution
- \(t = x^4\), \(dt = 4x^3\,dx\): \(\tfrac14\int\dfrac{dt}{\sqrt{1 - t^2}}\).
Answer: \(\tfrac14\sin^{-1}(x^4) + C\)
Practise this: Step 1, Secure the basics →
Miscellaneous Exercise, Question 13
\(\displaystyle\int \dfrac{e^x}{(1 + e^x)(2 + e^x)}\,dx\)
Show solution
- \(t = e^x\): \[\int\dfrac{dt}{(1 + t)(2 + t)} = \int\left(\dfrac{1}{1 + t} - \dfrac{1}{2 + t}\right)dt\]
Answer: \(\log\dfrac{1 + e^x}{2 + e^x} + C\)
Practise this: Step 2, Board standard →
Miscellaneous Exercise, Question 14
\(\displaystyle\int \dfrac{1}{(x^2 + 1)(x^2 + 4)}\,dx\)
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- \[\dfrac{1}{(x^2 + 1)(x^2 + 4)} = \dfrac13\left[\dfrac{1}{x^2 + 1} - \dfrac{1}{x^2 + 4}\right]\]
Answer: \[\tfrac13\tan^{-1}x - \tfrac16\tan^{-1}\dfrac{x}{2} + C\]
Practise this: Step 2, Board standard →
Miscellaneous Exercise, Question 15
\(\displaystyle\int \cos^3 x\,e^{\log\sin x}\,dx\)
Show solution
- \(e^{\log\sin x} = \sin x\); \(t = \cos x\): \(-\int t^3\,dt\).
Answer: \(-\tfrac14\cos^4 x + C\)
Practise this: Step 1, Secure the basics →
Miscellaneous Exercise, Question 16
\(\displaystyle\int e^{3\log x}(x^4 + 1)^{-1}\,dx\)
Show solution
- \(e^{3\log x} = x^3\), so \(\dfrac{x^3}{x^4 + 1}\).
Answer: \(\tfrac14\log(x^4 + 1) + C\)
Practise this: Step 1, Secure the basics →
Miscellaneous Exercise, Question 17
\(\displaystyle\int f'(ax + b)[f(ax + b)]^n\,dx\)
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- \(t = f(ax + b)\), \(dt = af'(ax + b)\,dx\): \(\dfrac1a\int t^n\,dt\) (\(n \ne -1\)).
Answer: \(\dfrac{[f(ax + b)]^{n+1}}{a(n + 1)} + C\)
Practise this: Step 3, Full marks on long answers →
Miscellaneous Exercise, Question 18
\(\displaystyle\int \dfrac{1}{\sqrt{\sin^3 x\sin(x + \alpha)}}\,dx\)
Show solution
- \[\sin^3 x\sin(x + \alpha) = \sin^4 x(\cos\alpha + \cot x\sin\alpha)\], so the integrand is \[\dfrac{\csc^2 x}{\sqrt{\cos\alpha + \cot x\sin\alpha}}\]
- \(t = \cos\alpha + \cot x\sin\alpha\), \(dt = -\csc^2 x\sin\alpha\,dx\): \[-\dfrac{1}{\sin\alpha}\int t^{-1/2}\,dt = -\dfrac{2}{\sin\alpha}\sqrt t\], and \(t = \dfrac{\sin(x + \alpha)}{\sin x}\).
Answer: \[-\dfrac{2}{\sin\alpha}\sqrt{\dfrac{\sin(x + \alpha)}{\sin x}} + C\]
Practise this: Step 4, 95+ stretch (HOTS) →
Miscellaneous Exercise, Question 19
\(\displaystyle\int \sqrt{\dfrac{1 - \sqrt x}{1 + \sqrt x}}\,dx\)
Show solution
- \(x = \cos^2\theta\), \(dx = -2\sin\theta\cos\theta\,d\theta\); \[\sqrt{\dfrac{1 - \cos\theta}{1 + \cos\theta}} = \tan\dfrac{\theta}{2}\]
- \[\begin{aligned}\tan\dfrac{\theta}{2} \cdot 2\sin\theta &= 4\sin^2\dfrac{\theta}{2} \\ &= 2(1 - \cos\theta)\end{aligned}\], so the integral is \[-\int 2\cos\theta(1 - \cos\theta)\,d\theta = -2\sin\theta + \theta + \tfrac12\sin 2\theta\]
- Back to x: \(\sin\theta = \sqrt{1 - x}\), \(\cos\theta = \sqrt x\), \(\theta = \cos^{-1}\sqrt x\).
Answer: \[-2\sqrt{1 - x} + \cos^{-1}\sqrt x + \sqrt{x(1 - x)} + C\]
Practise this: Step 4, 95+ stretch (HOTS) →
Miscellaneous Exercise, Question 20
\(\displaystyle\int \dfrac{2 + \sin 2x}{1 + \cos 2x}e^x\,dx\)
Show solution
- \[\dfrac{2 + 2\sin x\cos x}{2\cos^2 x} = \sec^2 x + \tan x\]: form \(e^x[f + f']\) with \(f = \tan x\).
Answer: \(e^x\tan x + C\)
Practise this: Step 3, Full marks on long answers →
Miscellaneous Exercise, Question 21
\(\displaystyle\int \dfrac{x^2 + x + 1}{(x + 1)^2(x + 2)}\,dx\)
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- \[\dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2} + \dfrac{C}{x + 2}\]: \(C = 3\), \(B = 1\), and \(A + C = 1\) gives \(A = -2\).
Answer: \[-2\log|x + 1| - \dfrac{1}{x + 1} + 3\log|x + 2| + C\]
Practise this: Step 2, Board standard →
Miscellaneous Exercise, Question 22
\(\displaystyle\int \tan^{-1}\sqrt{\dfrac{1 - x}{1 + x}}\,dx\)
Show solution
- \(x = \cos\theta\): \[\sqrt{\dfrac{1 - \cos\theta}{1 + \cos\theta}} = \tan\dfrac{\theta}{2}\], so the integrand is \(\dfrac{\theta}{2} = \tfrac12\cos^{-1}x\).
- \[\int\cos^{-1}x\,dx = x\cos^{-1}x - \sqrt{1 - x^2}\] (parts).
Answer: \[\tfrac12\left(x\cos^{-1}x - \sqrt{1 - x^2}\right) + C\]
Practise this: Step 3, Full marks on long answers →
Miscellaneous Exercise, Question 23
\(\displaystyle\int \dfrac{\sqrt{x^2 + 1}\,[\log(x^2 + 1) - 2\log x]}{x^4}\,dx\)
Show solution
- \[\log(x^2 + 1) - 2\log x = \log\left(1 + \dfrac{1}{x^2}\right)\] and \[\dfrac{\sqrt{x^2 + 1}}{x^4} = \dfrac{1}{x^3}\sqrt{1 + \dfrac{1}{x^2}}\]
- \(t = 1 + \dfrac{1}{x^2}\), \(dt = -\dfrac{2}{x^3}\,dx\): \[-\tfrac12\int t^{1/2}\log t\,dt = -\tfrac12\left[\tfrac23t^{3/2}\log t - \tfrac49t^{3/2}\right]\]
Answer: \[-\tfrac13\left(1 + \dfrac{1}{x^2}\right)^{3/2}\left[\log\left(1 + \dfrac{1}{x^2}\right) - \dfrac23\right] + C\]
Practise this: Step 4, 95+ stretch (HOTS) →
Miscellaneous Exercise, Question 24
\(\displaystyle\int_{\frac{\pi}{2}}^{\pi} e^x\left(\dfrac{1 - \sin x}{1 - \cos x}\right)\,dx\)
Show solution
- \[\dfrac{1 - \sin x}{1 - \cos x} = \tfrac12\csc^2\dfrac{x}{2} - \cot\dfrac{x}{2}\], and \[\dfrac{d}{dx}\left(-e^x\cot\dfrac{x}{2}\right) = e^x\left(\tfrac12\csc^2\dfrac{x}{2} - \cot\dfrac{x}{2}\right)\]
- \[\left[-e^x\cot\dfrac{x}{2}\right]_{\pi/2}^{\pi} = 0 + e^{\pi/2}\]
Answer: \(e^{\pi/2}\)
Practise this: Step 3, Full marks on long answers →
Miscellaneous Exercise, Question 25
\(\displaystyle\int_{0}^{\frac{\pi}{4}} \dfrac{\sin x\cos x}{\cos^4 x + \sin^4 x}\,dx\)
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- Divide by \(\cos^4 x\): \(\dfrac{\tan x\sec^2 x}{1 + \tan^4 x}\); \(t = \tan^2 x\), \(dt = 2\tan x\sec^2 x\,dx\).
- \[\tfrac12\int_0^1\dfrac{dt}{1 + t^2} = \tfrac12 \cdot \tfrac{\pi}{4}\]
Answer: \(\tfrac{\pi}{8}\)
Practise this: Step 3, Full marks on long answers →
Miscellaneous Exercise, Question 26
\(\displaystyle\int_{0}^{\frac{\pi}{2}} \dfrac{\cos^2 x}{\cos^2 x + 4\sin^2 x}\,dx\)
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- \[\dfrac{\cos^2 x}{\cos^2 x + 4\sin^2 x} = \dfrac{1}{1 + 4\tan^2 x}\]; \(t = \tan x\), \(dx = \dfrac{dt}{1 + t^2}\), t from 0 to \(\infty\).
- \[\dfrac{1}{(1 + t^2)(1 + 4t^2)} = \dfrac13\left[\dfrac{4}{1 + 4t^2} - \dfrac{1}{1 + t^2}\right]\]: \[\tfrac13\left[2 \cdot \tfrac{\pi}{2} - \tfrac{\pi}{2}\right]\]
Answer: \(\tfrac{\pi}{6}\)
Practise this: Step 4, 95+ stretch (HOTS) →
Miscellaneous Exercise, Question 27
\(\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{\sin x + \cos x}{\sqrt{\sin 2x}}\,dx\)
Show solution
- \(t = \sin x - \cos x\), \(dt = (\cos x + \sin x)\,dx\), \(\sin 2x = 1 - t^2\); t runs from \(\tfrac{1 - \sqrt3}{2}\) to \(\tfrac{\sqrt3 - 1}{2}\).
- \([\sin^{-1}t]\) between these limits, an odd function: twice the upper value.
Answer: \(2\sin^{-1}\dfrac{\sqrt3 - 1}{2}\)
Practise this: Step 4, 95+ stretch (HOTS) →
Miscellaneous Exercise, Question 28
\(\displaystyle\int_{0}^{1} \dfrac{1}{\sqrt{1 + x} - \sqrt x}\,dx\)
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- Rationalise: the integrand is \(\sqrt{1 + x} + \sqrt x\).
- \[\left[\tfrac23(1 + x)^{3/2} + \tfrac23x^{3/2}\right]_0^1 = \tfrac23(2\sqrt2 + 1) - \tfrac23\]
Answer: \(\dfrac{4\sqrt2}{3}\)
Practise this: Step 2, Board standard →
Miscellaneous Exercise, Question 29
\(\displaystyle\int_{0}^{\frac{\pi}{4}} \dfrac{\sin x + \cos x}{9 + 16\sin 2x}\,dx\)
Show solution
- \(t = \sin x - \cos x\) from \(-1\) to 0; \(\sin 2x = 1 - t^2\), so \(9 + 16\sin 2x = 25 - 16t^2\).
- \[\begin{aligned}\int_{-1}^{0}\dfrac{dt}{25 - 16t^2} &= \dfrac{1}{40}\left[\log\left|\dfrac{5 + 4t}{5 - 4t}\right|\right]_{-1}^{0} \\ &= \dfrac{1}{40}(0 - \log\tfrac19)\end{aligned}\]
Answer: \(\tfrac{1}{20}\log 3\)
Practise this: Step 4, 95+ stretch (HOTS) →
Miscellaneous Exercise, Question 30
\(\displaystyle\int_{0}^{\frac{\pi}{2}} \sin 2x\tan^{-1}(\sin x)\,dx\)
Show solution
- \(t = \sin x\): \(\int_0^1 2t\tan^{-1}t\,dt\).
- Parts: \[[t^2\tan^{-1}t]_0^1 - \int_0^1\dfrac{t^2}{1 + t^2}\,dt = \tfrac{\pi}{4} - \left(1 - \tfrac{\pi}{4}\right)\]
Answer: \(\dfrac{\pi}{2} - 1\)
Practise this: Step 3, Full marks on long answers →
Miscellaneous Exercise, Question 31
\(\displaystyle\int_{1}^{4} [|x - 1| + |x - 2| + |x - 3|]\,dx\)
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- \(\int_1^4|x - 1|\,dx = \tfrac92\); \[\begin{aligned}\int_1^4|x - 2|\,dx &= \tfrac12 + 2 \\ &= \tfrac52\end{aligned}\]; \[\begin{aligned}\int_1^4|x - 3|\,dx &= 2 + \tfrac12 \\ &= \tfrac52\end{aligned}\]
Answer: \(\tfrac{19}{2}\)
Practise this: Step 2, Board standard →
Miscellaneous Exercise, Question 32
Prove:
\(\displaystyle\int_1^3\dfrac{dx}{x^2(x + 1)} = \dfrac23 + \log\dfrac23\)
Show solution
- \[\dfrac{1}{x^2(x + 1)} = -\dfrac1x + \dfrac{1}{x^2} + \dfrac{1}{x + 1}\]
- \[\begin{aligned}\left[\log\dfrac{x + 1}{x} - \dfrac1x\right]_1^3 &= \left(\log\tfrac43 - \tfrac13\right) - (\log 2 - 1) \\ &= \log\tfrac23 + \tfrac23\end{aligned}\]
Answer: Proved
Practise this: Step 3, Full marks on long answers →
Miscellaneous Exercise, Question 33
\(\displaystyle\int_0^1 xe^x\,dx = 1\)
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- \([e^x(x - 1)]_0^1 = 0 - (-1)\).
Answer: Proved
Practise this: Step 2, Board standard →
Miscellaneous Exercise, Question 34
\(\displaystyle\int_{-1}^{1}x^{17}\cos^4 x\,dx = 0\)
Show solution
- \(x^{17}\cos^4 x\) is odd.
Answer: Proved
Practise this: Step 2, Board standard →
Miscellaneous Exercise, Question 35
\(\displaystyle\int_0^{\pi/2}\sin^3 x\,dx = \dfrac23\)
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- \(\sin^3 x = (1 - \cos^2 x)\sin x\); \(t = \cos x\) from 1 to 0: \(\int_0^1(1 - t^2)\,dt\).
Answer: Proved
Practise this: Step 2, Board standard →
Miscellaneous Exercise, Question 36
\(\displaystyle\int_0^{\pi/4}2\tan^3 x\,dx = 1 - \log 2\)
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- \(2\tan x(\sec^2 x - 1)\): \[\left[\tan^2 x - 2\log|\sec x|\right]_0^{\pi/4} = 1 - 2\log\sqrt2\]
Answer: Proved
Practise this: Step 2, Board standard →
Miscellaneous Exercise, Question 37
\(\displaystyle\int_0^1\sin^{-1}x\,dx = \dfrac{\pi}{2} - 1\)
Show solution
- Parts: \[\left[x\sin^{-1}x + \sqrt{1 - x^2}\right]_0^1 = \tfrac{\pi}{2} - 1\]
Answer: Proved
Practise this: Step 2, Board standard →
Miscellaneous Exercise, Question 38
\(\displaystyle\int\dfrac{dx}{e^x + e^{-x}}\) is: (A) \(\tan^{-1}(e^x) + C\) (B) \(\tan^{-1}(e^{-x}) + C\) (C) \(\log(e^x - e^{-x}) + C\) (D) \(\log(e^x + e^{-x}) + C\)
Show solution
- \(\dfrac{e^x}{e^{2x} + 1}\); \(t = e^x\): \(\int\dfrac{dt}{1 + t^2}\).
Answer: (A)
Practise this: Step 2, Board standard →
Miscellaneous Exercise, Question 39
\(\displaystyle\int\dfrac{\cos 2x}{(\sin x + \cos x)^2}\,dx\) is: (A) \(\dfrac{-1}{\sin x + \cos x} + C\) (B) \(\log|\sin x + \cos x| + C\) (C) \(\log|\sin x - \cos x| + C\) (D) \(\dfrac{1}{(\sin x + \cos x)^2}\)
Show solution
- \[\dfrac{\cos 2x}{(\sin x + \cos x)^2} = \dfrac{\cos x - \sin x}{\cos x + \sin x}\]
Answer: (B)
Practise this: Step 2, Board standard →
Miscellaneous Exercise, Question 40
If \(f(a + b - x) = f(x)\), then \(\displaystyle\int_a^b xf(x)\,dx\) is: (A) \(\dfrac{a + b}{2}\int_a^b f(b - x)\,dx\) (B) \(\dfrac{a + b}{2}\int_a^b f(b + x)\,dx\) (C) \(\dfrac{b - a}{2}\int_a^b f(x)\,dx\) (D) \(\dfrac{a + b}{2}\int_a^b f(x)\,dx\)
Show solution
- \[\begin{aligned}I &= \int_a^b(a + b - x)f(a + b - x)\,dx \\ &= \int_a^b(a + b - x)f(x)\,dx \\ &= (a + b)\int_a^b f(x)\,dx - I\end{aligned}\]
Answer: (D)
Practise this: Step 3, Full marks on long answers →
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