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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} $

1 Answer

  • (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

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