Question

Evaluate $\int \frac{\sin x}{({\cos }^{2}x+1)({\cos }^{2}x+4)}dx$.

Answer 1

$I=\int \frac{\sin x}{({\cos }^{2}x+1)({\cos }^{2}x+4)}dx$

Put $\cos x=t⟹-\sin xdx=dt$

Therefore,

$I=\int \frac{-dt}{(t^2+1)(t^2+4)}$ .......(1)

Again, consider $t^2=y$

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

$-1=A(y+4)+B(y+1)$

$-1=(A+B)y+4A+B$

On equating the coefficient of y and constant term both sides, we get,

$0=A+B$

and $-1=4A+B$

$⟹A=-\frac{1}{3}$ and $B=\frac{1}{3}$

Therefore,

$-\frac{1}{(y+1)(y+4)}=-\frac{1}{3(y+1)}+\frac{1}{3(y+4)}$

$-\frac{1}{(t^2+1)(t^2+4)}=-\frac{1}{3(t^2+1)}+\frac{1}{3(t^2+4)}$

From equation 1, we get,

$I=\int -\frac{1}{3(t^2+1)}dt+\int \frac{1}{3(t^2+4)}dt$

$I=-\frac{1}{3}\times {\tan }^{-1}t+\frac{1}{3}\times \frac{1}{2}{\tan }^{-1}\frac{t}{2}+C$ ............. $\int \frac{1}{a^2+x^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+c$

$I=-\frac{1}{3}{\tan }^{-1}t+\frac{1}{6}{\tan }^{-1}\frac{t}{2}+C$

$I=-\frac{1}{3}{\tan }^{-1}\left(\cos x\right)+\frac{1}{6}{\tan }^{-1}\left(\frac{\cos x}{2}\right)+C$