Question

The value of the integral $\int \frac{{\cos }^{3}x+{\cos }^{5}x}{{\sin }^{2}x+{\sin }^{4}x}dx$ is

• A. $\sin x-6{\tan }^{-1}\left(\sin x\right)-2\left(si{n}^{-1}x\right)+c$
• B. $\sin x-2{\left(\sin x\right)}^{-1}+c$
• C. $\sin x-2{\left(\sin x\right)}^{-1}-6{\tan }^{-1}\left(\sin x\right)+c$
• D. $\sin x-2{\left(\sin x\right)}^{-1}+5{\tan }^{-1}\left(\sin x\right)+c$
• E. $Noneofthese$

Answer 1

The correct option is A $\sin x-6{\tan }^{-1}\left(\sin x\right)-2\left(si{n}^{-1}x\right)+c$

$\int \frac{{\cos }^{3}x+{\cos }^{5}x}{{\sin }^{2}x+{\sin }^{4}x}dx$

$\because {\cos }^{2}x=1-{\sin }^{2}x$

$=\int \cos x\frac{{\sin }^{4}x-3{\sin }^x+2}{{\sin }^{2}x({\sin }^{2}x+1}dx$

Let $u=\sin x$ so, $dx=\frac{1}{\cos x}du$

$=\int \frac{2-4u^2}{u^2(u^2+1)}+1du$

$=\int 1du-2\int \frac{2u^2-1}{u^2(u^2+1)}du$

$=u-6\int \frac{1}{u^2+1}+2\int \frac{1}{u^2}du+c$

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

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

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

$I=\sin x-6{\tan }^{-1}\left(\sin x\right)-2\left(si{n}^{-1}x\right)+c$