Indefinite Integration 1 Question 14

15. The value of $\int _0^{\pi}|\cos x|^{3} d x$ is

(2019 Main, 9 Jan I)

(a) $\frac{2}{3}$

(b) $-\frac{4}{3}$

(c) 0

(d) $\frac{4}{3}$

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Solution:

  1. We know, graph of $y=\cos x$ is

$\therefore$ The graph of $y=|\cos x|$ is

$$ I=\int _0^{\pi}|\cos x|^{3}=2 \int _0^{\frac{\pi}{2}}|\cos x|^{3} d x $$

$\left(\because y=|\cos x|\right.$ is symmetric about $\left.x=\frac{\pi}{2}\right)$

$$ =2 \int _0^{\frac{\pi}{2}} \cos ^{3} x d x \quad \because \cos x \geq 0 \text { for } x \in 0, \frac{\pi}{2} $$

Now, as $\cos 3 x=4 \cos ^{3} x-3 \cos x$

$\therefore \cos ^{3} x=\frac{1}{4}(\cos 3 x+3 \cos x)$

$\therefore I=\frac{2}{4} \int _0^{\frac{\pi}{2}}(\cos 3 x+3 \cos x) d x$

$=\frac{1}{2} \frac{\sin 3 x}{3}+3 \sin x _0^{\frac{\pi}{2}}$

$=\frac{1}{2} \quad \frac{1}{3} \sin \frac{3 \pi}{2}+3 \sin \frac{\pi}{2}-\frac{1}{3} \sin 0+3 \sin 0$

$=\frac{1}{2} \quad \frac{1}{3}(-1)+3-[0+0]$

$\because \sin \frac{3 \pi}{2}=\sin \pi+\frac{\pi}{2}=-\sin \frac{\pi}{2}=-1$

$=\frac{1}{2}-\frac{1}{3}+3=\frac{4}{3}$



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