Question

Evaluate the integrals: $\int^\frac{\pi}{2}_0 {\frac{sin\ x}{1+ cos^2 \x}$

Answer 1

Answer:

Step-by-step explanation:

Concept:

I have applied change of variable method to find the integral of the given function.

Formula used:

$\int{\frac{1}{1+x^2}}\:dx=tan^{-1}x+c$

Let

$I=\int\limits^{\frac{\pi}{2}}_0{\frac{sinx}{1+cos^2x}}\:dx$

Take,

$t=cosx\\\\\frac{dt}{dx}=-sinx\\\\-dt=sinx\:dx$

$when\:x=0, t=cos0=1\\\\when\:x=\frac{\pi}{2}, t=cos\frac{\pi}{2}=0\\\\Now,\\\\I=\int\limits^{0}_1{\frac{1}{1+t^2}}\:(-dt)\\\\I=-\int\limits^{0}_1{\frac{1}{1+t^2}}\:dt\\\\I=-[tan^{-1}t]^{0}_1\\\\I=-[tan^{-1}0-tan^{-1}1]\\\\I=-[0-\frac{\pi}{4}]\\\\I=\frac{\pi}{4}$