Browse Questions

Evaluate:$\int\limits_{-\pi/2}^ {\pi/2} |\sin x|dx$

Toolbox:
• $(i)\; \int \limits_a^b f(x)dx=F(b)-F(a)$
• $(ii)\;\int \limits_{-a}^a f(x) dx=2 \int \limits_0^a f(x) dx$ is an even function. If $f(-x)=+f(x)$ then $f(x)$ is an even function
Step1:
Given $I=\int \limits_{-\pi/2}^{\pi/2} |\sin x| dx$
Let $f(x) =|\sin x|$
Therefore $f(-x)=|\sin (-x)|=|-\sin x|=\sin x =f(x)$
Therefore $|\sin x|$ is an even function
Hence $\int \limits_{-a}^a f(x) dx=2 \int \limits_0^{a} f(x)dx$ if it is an even function
Therefore $I=2 \int \limits_{-\pi/2}^{\pi/2}|\sin x| dx=2\int \limits_0^{\pi/2} |\sin x|dx$
Therefore $I=2 \int \limits_0^{\pi/2} \sin x dx$
This is because $\sin x$ is positive on $[0,\pi/2]$ and hence $|\sin x|=\sin x$
Step 2:
On integrating we get,
$2 \bigg[\cos x\bigg]_0^{\pi/2}=2$
On applying limits we get
$2 [\cos \pi/2-\cos 0]$
But $\cos \pi/2=0 \;and\; \cos 0=1$
Hence $I=2(1) =2$
Therefore $\int \limits_{-\pi/2}^{\pi/2} |\sin x| dx=2$