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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

Integrate : $\int \limits_0^{\large\frac{\pi}{2}} \large\frac{\cos ^3 x}{\sin ^3 x + \cos ^3 x }$$ dx$

$ (A)\;\large\frac{\pi}{2} \\ (B)\;\large\frac{\pi}{4} \\ (C)\;2 \pi \\ (D)\;\large\frac{\pi}{6} $

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1 Answer

$f(x) = \int \limits_0^{\large\frac{\pi}{2}} \large\frac{\cos ^3 x}{\sin ^3 x + \cos ^3 x }$$ dx$... (1)
By using properties $f(x)= f( \large\frac{\pi}{2}$$-x)$
$f(\large\frac{\pi}{2}$$ - x)=>$ $\int \limits_0^{\large\frac{\pi}{2}} \large\frac{\cos ^3 (\Large\frac{\pi}{2}-x)}{\sin^3 (\large\frac{\pi}{2}-x) + \cos ^3 (\Large\frac{\pi}{2}-x)}$$dx$
$f(\large\frac{\pi}{2}$$-0)=>$ $\int \limits_0^{\large\frac{\pi}{2}} \large\frac{\sin ^3 x}{\cos ^3 x +\sin ^3 x}$----(2)
adding (1) and (2) we get,
$f(x) +f(\large\frac{\pi }{2} $$-x)=\int \limits_0^{\large\frac{\pi}{2}}$$ dx$
$2f(x)=\large\frac{z}{2}$$ [1-0]$
$f(x)= \large\frac{\pi}{4}$
Hence b is the correct answer.
answered Dec 13, 2013 by meena.p
edited Mar 20, 2014 by balaji.thirumalai
 
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