Question

Find $\int \frac{sinx}{(co{s}^{2}x+1)(co{s}^{2}x+4)}dx.$

Answer 1

Let $I=\int \frac{sinx}{(co{s}^{2}x+1)(co{s}^{2}x+4)}dx$

Put $cosx=t\Rightarrow sinxdx=-dt$ so, $I=\int \frac{-dt}{(t^2+1)(t^2+4)}$

Consider $\frac{-1}{(t^2+1)(t^2+4)}=\frac{-1}{(1+y)(4+y)}=\frac{A}{1+y}+\frac{B}{4+y}$ where $y=t^2$

$\Rightarrow -1=A(4+y)+B(1+y)\therefore A=-\frac{1}{3},B=\frac{1}{3}$

Therefore, $I=\int \frac{1}{3}[\frac{1}{t^2+4}-\frac{1}{t^2+1}]dt\Rightarrow I=\frac{1}{3}[\frac{1}{2}ta{n}^{-1}\frac{t}{2}-ta{n}^{-1}t]+C$

$\therefore I=\frac{1}{6}ta{n}^{-1}\left(\frac{cosx}{2}\right)-\frac{1}{3}ta{n}^{-1}\left(cosx\right)+C.$