IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

ramkumar_november

Here is my solution.........

I = $\int_0^\frac{\pi}{2}\frac{1+2cosx}{(2+cosx)^2}\;dx$

$divide\;numerator\;and\;denominator\;by\;sin^2x$ .......

I = $\int_0^\frac{\pi}{2}\frac{cosec^2x+2cotxcosecx}{(2cosecx+cotx)^2}\;dx$

$now\;put\;2cosecx+cotx\;=\;t$

differentiating we get.......

$-(cosec^2x+2cosecxcotx)dx\;=\;dt$

so the integral becomes

I = $\int_2^\infty\frac{1}{t^2}\;dt$

I = $\Bigg[\frac{-1}{t}\Bigg]_2^\infty$

I = $\frac{1}{2}$