Question

Evaluate the integral using substitution:
${\int }_0^{\frac{\pi }{2}}\frac{\sin x}{1+{\cos }^{2}x}dx$

Answer 1

Evaluate the integral using substitution:
${\int }_0^{\frac{\pi }{2}}\frac{\sin x}{1+{\cos }^{2}x}dx$

Let $\cos x=t$

Differentiate with respect to $t$

$=\frac{dt}{dx}=-\sin x\Rightarrow -dt=\sin xdx$

The new limits are

When

$x=0\Rightarrow t=\cos 0=1$

When

$x=2\Rightarrow t=\cos \frac{\pi }{2}=0$

Now,

${\int }_0^{\frac{\pi }{2}}\frac{\sin x}{1+{\cos }^{2}x}dx$

$={\int }_1^0\frac{-dt}{1+t^2}$

$=-{\left[{\tan }^{-1}t\right]}_1^0$

$=-[ta{n}^{-1}0-{\tan }^{-1}1]$

$=-[0-\frac{\pi }{4}]$

$=\frac{\pi }{4}$