# By using the properties of definite integrals,evaluate the integral$\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2}\sin^2\;x\;dx$

$\begin{array}{1 1} \frac{\pi}{2} \\ \frac{\pi}{4} \\ \frac{\pi}{16}\\ \frac{\pi}{8} \end{array}$

Toolbox:
• (i)$\int\limits_a^b f(x)dx=F(b)-F(a)$
• (ii)$If \;f(-x)=-f(x)$ then it is an odd function
• (iii)$If\;f(-x)=f(x),$ then it is an even function.
• (iv) if the given function is an even function, then $\int \limits_{-a}^a f(x)dx=2\int \limits_0^af(x)dx$
• (v) if the given function is an odd function, then $\int \limits_{-a}^a f(x)dx=0$

Given $I=\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2}\sin^2\;x\;dx$

$f(x)=\sin ^2 x;f(-x)=(-\sin x)^2=\sin^2x$

Hence the given function is an even function.

We know $\sin ^2x=\frac{1-\cos2x}{2}$

Therefore $\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2}\bigg(\frac{1-\cos 2x}{2}\bigg)dx$

Since it is an even function

$I=2 \times 2 \frac{1}{2} \int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2}(1-\cos 2x)dx$

$=\int \limits_0^{\frac{\pi}{2}} dx-\int \limits_0^{\frac{\pi}{2}}\cos 2x dx$

On integrating we get

$I=[x]_0^{\pi/2}-[\frac{\sin 2x}{2}]_0^{\pi/2}$

On applying limits we get,

$I=[\frac{\pi}{2}-0]-\frac{\pi}{2}[\sin 2.\frac{\pi}{2}-\sin 0]$

But $\sin \pi=\sin 0=0;$

Hence $I=\frac{\pi}{2}$