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) $\frac{4}{3}$
Show Answer
Solution:
- 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 $\cos 3 x=4 \cos ^{3} x-3 \cos x$
$\therefore \cos ^{3} x=\frac{1}{4}(\cos 3 x+3 \cos x)$
$\therefore I=\frac{2}{4} \int _0^{\frac{\pi}{2}}(\cos 3 x+3 \cos x) d x$
$=\frac{1}{2} \frac{\sin 3 x}{3}+3 \sin x _0^{\frac{\pi}{2}}$
$=\frac{1}{2} \quad \frac{1}{3} \sin \frac{3 \pi}{2}+3 \sin \frac{\pi}{2}-\frac{1}{3} \sin 0+3 \sin 0$
$=\frac{1}{2} \quad \frac{1}{3}(-1)+3-[0+0]$
$\because \sin \frac{3 \pi}{2}=\sin \pi+\frac{\pi}{2}=-\sin \frac{\pi}{2}=-1$
$=\frac{1}{2}-\frac{1}{3}+3=\frac{4}{3}$
