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Home  >>  CBSE XII  >>  Math  >>  Integrals
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By using the properties of definite integrals,evaluate the integral\[\int\limits_0^\frac{\Large \pi}{\Large 2}\cos^2x\;dx\]

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

Can you answer this question?
 
 

1 Answer

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Toolbox:
  • (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii)$\sin ^2x+\cos ^2x=1$
Given $\int\limits_0^\frac{\Large \pi}{\Large 2}\cos^2x\;dx-----(1)$
 
By applying the properties $\int \limits_0^a f(x)dx=\int _0^af(a-x)dx$
 
$I=\int\limits_0^\frac{\Large \pi}{\Large 2}\cos^2(\frac{\pi}{2}-x)\;dx$
 
But $\cos^2(\frac{\pi}{2}-x)=\sin^2x$
 
$I=\int \limits_0^{\frac{\pi}{2}} \sin ^2 xdx-----(2)$
 
Adding equation (1) and(2)
 
$2I=\int \limits_0^{\frac{\pi}{2}}( \sin ^2 x+\cos ^2 x)dx$
 
But $\sin^2x+\cos^2x=1$
 
Therefore $ 2I=\int \limits_0^{\frac{\pi}{2}}dx$
 
On integrating we get,
 
$2I=[x]_0^{\frac{\pi}{2}}$
 
On applying limits,
 
$2I=\frac{\pi}{2}-0$
 
$=\frac{\pi}{2}$
 
Therefore $=\frac{\pi}{4}$

 

 

answered Feb 13, 2013 by meena.p
 

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