Question

Evaluate the integrals using substitution.
${\int }_0^{\frac{\pi }{2}}\frac{sinx}{1+co{s}^{2}x}dx.$

Answer 1

Let $I={\int }_0^{\frac{\pi }{2}}\frac{sinx}{1+co{s}^{2}x}dx$
Put $cosx=t\Rightarrow -sinxdx=dt\Rightarrow dx=\frac{dt}{-sinx}$
For limit when $x=0\Rightarrow t=cos0=1(\because t=cosx)$ and when $x=\frac{\pi }{2}\Rightarrow t=cos\frac{\pi }{2}=0$
$\therefore I={\int }_1^0\frac{sinx}{1+t^2}\frac{dt}{(-sinx)}dx=-{\int }_1^0\frac{1}{1+t^2}dt(\because \int \frac{1}{a^2+x^2}dx=\frac{1}{a}ta{n}^{-1}\frac{x}{a})=-{\left[\frac{1}{1}ta{n}^{-1}\right(\frac{t}{1}\left)\right]}_1^0=-[ta{n}^{-1}t{]}_1^0=-[ta{n}^{-1}\left(0\right)-ta{n}^{-1}\left(1\right)]=-(0-\frac{\pi }{4})=\frac{\pi }{4}.$