Question

Integrate the following, $\int \frac{\sin x}{(1+{\cos }^{2}x)(4+{\cos }^{2}x)}$

• A. $\frac{-1}{3}{\tan }^{-1}\left(\cos x\right)+\frac{1}{6}{\tan }^{-1}\left(\frac{\cos x}{2}\right)+c$
• B. $\frac{{\tan }^{-1}\left(\frac{\cos x}{\sqrt{2}}\right)}{\sqrt{2}}+{\tan }^{-1}\left(\cos x\right)$
• C. $-\frac{{\tan }^{-1}\left(\frac{\cos x}{\sqrt{2}}\right)}{\sqrt{2}}+{\tan }^{-1}\left(\cos x\right)$
• D. None of these

Answer 1

The correct option is A $\frac{-1}{3}{\tan }^{-1}\left(\cos x\right)+\frac{1}{6}{\tan }^{-1}\left(\frac{\cos x}{2}\right)+c$

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

putting $\cos x=t$

$\Rightarrow -\sin xdx=dt$

$\Rightarrow \sin xdx=-dt$

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

putting $t^2=y$

$={\int }^{}\frac{-dt}{\left(y+1\right)\left(y+4\right)}$

Now, putting $\frac{-1}{\left(y+1\right)\left(y+4\right)}=\frac{A}{y+1}+\frac{B}{y+4}$

We will get

$A=\frac{-1}{3}$ and $\text{B =}\frac{1}{3}$

$={\int }^{}\left(\frac{\frac{-1}{3}}{y+1}+\frac{\frac{1}{3}}{y+4}\right)dt$

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

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

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