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NCERT Solutions · Class 12 · Chapter 7: Integrals
NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10
Exercise 7.10: Properties of definite integrals. \(\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx\) (add the two forms of I); \(\int_{-a}^a f = 2\int_0^a f\) for even f and 0 for odd f; \(\int_0^{2a} f = 2\int_0^a f\) if \(f(2a - x) = f(x)\), 0 if \(f(2a - x) = -f(x)\); split at the zeros of \(|\,\cdot\,|\) expressions.
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Exercise 7.10 questions and solutions
Exercise 7.10, Question 1
\(\displaystyle\int_{0}^{\frac{\pi}{2}} \cos^2 x\,dx\)
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With \(P_4\): \(I = \int_0^{\pi/2}\sin^2 x\,dx\), so \[\begin{aligned}2I &= \int_0^{\pi/2}1\,dx \\ &= \tfrac{\pi}{2}\end{aligned}\] Answer: \(\tfrac{\pi}{4}\)
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Exercise 7.10, Question 2
\(\displaystyle\int_{0}^{\frac{\pi}{2}} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx\)
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\(P_4\): \[\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx\] Call the integral I, write it again with \(x \to a - x\), and add. \(x \to \tfrac{\pi}{2} - x\) swaps sin and cos: \(2I = \int_0^{\pi/2}1\,dx\). Answer: \(\tfrac{\pi}{4}\)
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Exercise 7.10, Question 3
\(\displaystyle\int_{0}^{\frac{\pi}{2}} \dfrac{\sin^{3/2}x}{\sin^{3/2}x + \cos^{3/2}x}\,dx\)
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\(P_4\): \[\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx\] Call the integral I, write it again with \(x \to a - x\), and add. Adding the two forms gives \(2I = \tfrac{\pi}{2}\). Answer: \(\tfrac{\pi}{4}\)
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Exercise 7.10, Question 4
\(\displaystyle\int_{0}^{\frac{\pi}{2}} \dfrac{\cos^5 x}{\sin^5 x + \cos^5 x}\,dx\)
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\(P_4\): \[\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx\] Call the integral I, write it again with \(x \to a - x\), and add. \(2I = \tfrac{\pi}{2}\). Answer: \(\tfrac{\pi}{4}\)
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Exercise 7.10, Question 5
\(\displaystyle\int_{-5}^{5} |x + 2|\,dx\)
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Split at \(-2\): \[\int_{-5}^{-2}-(x + 2)\,dx + \int_{-2}^{5}(x + 2)\,dx = \tfrac92 + \tfrac{49}{2}\] Answer: \(29\)
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Exercise 7.10, Question 6
\(\displaystyle\int_{2}^{8} |x - 5|\,dx\)
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Split at 5: \[\int_2^5(5 - x)\,dx + \int_5^8(x - 5)\,dx = \tfrac92 + \tfrac92\] Answer: \(9\)
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Exercise 7.10, Question 7
\(\displaystyle\int_0^1 x(1 - x)^n\,dx\)
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\(P_4\): \[\begin{aligned}\int_0^1(1 - x)x^n\,dx &= \int_0^1(x^n - x^{n+1})\,dx \\ &= \dfrac{1}{n + 1} - \dfrac{1}{n + 2}\end{aligned}\] Answer: \(\dfrac{1}{(n + 1)(n + 2)}\)
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Exercise 7.10, Question 8
\(\displaystyle\int_{0}^{\frac{\pi}{4}} \log(1 + \tan x)\,dx\)
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\(x \to \tfrac{\pi}{4} - x\): \[\begin{aligned}1 + \tan\left(\tfrac{\pi}{4} - x\right) &= 1 + \dfrac{1 - \tan x}{1 + \tan x} \\ &= \dfrac{2}{1 + \tan x}\end{aligned}\] So \[\begin{aligned}I &= \int_0^{\pi/4}[\log 2 - \log(1 + \tan x)]\,dx \\ &= \tfrac{\pi}{4}\log 2 - I\end{aligned}\] Answer: \(\dfrac{\pi}{8}\log 2\)
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Exercise 7.10, Question 9
\(\displaystyle\int_{0}^{2} x\sqrt{2 - x}\,dx\)
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\(P_4\): \[\begin{aligned}\int_0^2(2 - x)\sqrt x\,dx &= \left[\tfrac43x^{3/2} - \tfrac25x^{5/2}\right]_0^2 \\ &= \tfrac43 \cdot 2\sqrt2 - \tfrac25 \cdot 4\sqrt2\end{aligned}\] Answer: \(\dfrac{16\sqrt2}{15}\)
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Exercise 7.10, Question 10
\(\displaystyle\int_{0}^{\frac{\pi}{2}} (2\log\sin x - \log\sin 2x)\,dx\)
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\[\log\sin 2x = \log 2 + \log\sin x + \log\cos x\], so the integrand is \(\log\sin x - \log\cos x - \log 2\). By \(P_4\), \[\int_0^{\pi/2}\log\sin x\,dx = \int_0^{\pi/2}\log\cos x\,dx\]; what is left is \(-\tfrac{\pi}{2}\log 2\). Answer: \(\dfrac{\pi}{2}\log\dfrac12\)
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Exercise 7.10, Question 11
\(\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x\,dx\)
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Even: \[2\int_0^{\pi/2}\sin^2 x\,dx = 2 \cdot \tfrac{\pi}{4}\] Answer: \(\tfrac{\pi}{2}\)
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Exercise 7.10, Question 12
\(\displaystyle\int_{0}^{\pi} \dfrac{x}{1 + \sin x}\,dx\)
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\(x \to \pi - x\) (\(\sin(\pi - x) = \sin x\)): \[I = \int_0^{\pi}\dfrac{\pi - x}{1 + \sin x}\,dx\]; adding, \[2I = \pi\int_0^{\pi}\dfrac{dx}{1 + \sin x}\] For \(x \ne \tfrac{\pi}{2}\): \[\begin{aligned}\dfrac{1}{1 + \sin x} &= \dfrac{1 - \sin x}{\cos^2 x} \\ &= \sec^2 x - \sec x\tan x\end{aligned}\], with antiderivative \[\begin{aligned}\tan x - \sec x &= \dfrac{\sin x - 1}{\cos x} \\ &= -\dfrac{\cos x}{1 + \sin x}\end{aligned}\] \(F(x) = -\dfrac{\cos x}{1 + \sin x}\) is defined on the whole of \([0, \pi]\) and \(F'(x) = \dfrac{1}{1 + \sin x}\) there, so \[\begin{aligned}\int_0^{\pi}\dfrac{dx}{1 + \sin x} &= F(\pi) - F(0) \\ &= 1 - (-1) \\ &= 2\end{aligned}\] and \(2I = 2\pi\). Answer: \(\pi\)
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Exercise 7.10, Question 13
\(\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^7 x\,dx\)
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\(\sin^7 x\) is odd. Answer: \(0\)
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Exercise 7.10, Question 14
\(\displaystyle\int_{0}^{2\pi} \cos^5 x\,dx\)
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\(f(2\pi - x) = \cos^5 x = f(x)\): \(I = 2\int_0^{\pi}\cos^5 x\,dx\); and \(\cos^5(\pi - x) = -\cos^5 x\), so \(\int_0^{\pi}\cos^5 x\,dx = 0\). Answer: \(0\)
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Exercise 7.10, Question 15
\(\displaystyle\int_{0}^{\frac{\pi}{2}} \dfrac{\sin x - \cos x}{1 + \sin x\cos x}\,dx\)
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\(x \to \tfrac{\pi}{2} - x\) turns the integrand into its negative, so \(I = -I\). Answer: \(0\)
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Exercise 7.10, Question 16
\(\displaystyle\int_{0}^{\pi} \log(1 + \cos x)\,dx\)
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\(x \to \pi - x\): \(I = \int_0^{\pi}\log(1 - \cos x)\,dx\). Add: \[\begin{aligned}2I &= \int_0^{\pi}\log\sin^2 x\,dx \\ &= 2\int_0^{\pi}\log\sin x\,dx\end{aligned}\] \[\int_0^{\pi}\log\sin x\,dx = 2\int_0^{\pi/2}\log\sin x\,dx\] by \(P_6\), as \(\sin(\pi - x) = \sin x\). Let \(J = \int_0^{\pi/2}\log\sin x\,dx\); by \(P_4\) also \(J = \int_0^{\pi/2}\log\cos x\,dx\). Add: \[\begin{aligned}2J &= \int_0^{\pi/2}\log(\sin x\cos x)\,dx \\ &= \int_0^{\pi/2}\log\sin 2x\,dx - \tfrac{\pi}{2}\log 2\end{aligned}\] With \(u = 2x\): \[\begin{aligned}\int_0^{\pi/2}\log\sin 2x\,dx &= \tfrac12\int_0^{\pi}\log\sin u\,du \\ &= \tfrac12 \cdot 2J \\ &= J\end{aligned}\] So \(2J = J - \tfrac{\pi}{2}\log 2\), i.e. \(J = -\tfrac{\pi}{2}\log 2\). \(I = 2J = -\pi\log 2\). Answer: \(-\pi\log 2\)
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Exercise 7.10, Question 17
\(\displaystyle\int_0^a\dfrac{\sqrt x}{\sqrt x + \sqrt{a - x}}\,dx\)
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\(P_4\): \[\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx\] Call the integral I, write it again with \(x \to a - x\), and add. \(2I = \int_0^a 1\,dx = a\). Answer: \(\dfrac{a}{2}\)
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Exercise 7.10, Question 18
\(\displaystyle\int_{0}^{4} |x - 1|\,dx\)
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\[\int_0^1(1 - x)\,dx + \int_1^4(x - 1)\,dx = \tfrac12 + \tfrac92\] Answer: \(5\)
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Exercise 7.10, Question 19
Show \(\displaystyle\int_0^a f(x)g(x)\,dx = 2\int_0^a f(x)\,dx\) if \(f(x) = f(a - x)\) and \(g(x) + g(a - x) = 4\).
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\[\begin{aligned}I &= \int_0^a f(a - x)g(a - x)\,dx \\ &= \int_0^a f(x)[4 - g(x)]\,dx \\ &= 4\int_0^a f(x)\,dx - I\end{aligned}\] So \(2I = 4\int_0^a f(x)\,dx\). Answer: Shown
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Exercise 7.10, Question 20
\(\displaystyle\int_{-\pi/2}^{\pi/2}(x^3 + x\cos x + \tan^5 x + 1)\,dx\) is: (A) 0 (B) 2 (C) \(\pi\) (D) 1
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\(x^3\), \(x\cos x\), \(\tan^5 x\) are odd and integrate to 0 over the symmetric interval; \(\int_{-\pi/2}^{\pi/2}1\,dx = \pi\). Answer: (C) \(\pi\)
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Exercise 7.10, Question 21
\(\displaystyle\int_0^{\pi/2}\log\left(\dfrac{4 + 3\sin x}{4 + 3\cos x}\right)dx\) is: (A) 2 (B) \(\tfrac34\) (C) 0 (D) \(-2\)
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\(x \to \tfrac{\pi}{2} - x\) turns the log into its negative: \(I = -I\). Answer: (C) 0
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