Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
  • (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 $I=\frac{\pi}{2}$


answered Feb 14, 2013 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App