Question

$\int \frac{{\sin }^{3}x+{\sin }^{5}x}{{\cos }^{2}x+{\cos }^{4}x}$ is equal to
(where $C$ is the constant of integration)

• A. $\sin x-2(\sin x{)}^{-1}+C$
• B. $\cos x-6{\tan }^{-1}\left(\cos x\right)+C$
• C. $\sin x-2(\sin x{)}^{-1}-6{\tan }^{-1}(\sin x)+C$
• D. $6{\tan }^{-1}\left(\cos x\right)+2(\cos x{)}^{-1}-\cos x+C$

Answer 1

The correct option is D $6{\tan }^{-1}\left(\cos x\right)+2(\cos x{)}^{-1}-\cos x+C$

Let $I=\int \frac{{\sin }^{3}x+{\sin }^{5}x}{{\cos }^{2}x+{\cos }^{4}x}$
$I=\int \frac{({\sin }^{2}x+{\sin }^{4}x)\sin x}{{\cos }^{2}x+{\cos }^{4}x}dx$
Put $\cos x=t\Rightarrow \sin xdx=-dt$
$I=-\int \frac{(1-t^2)+(1-t^2{)}^2}{t^2+t^4}dt$
$\Rightarrow I=-\int \frac{2-3t^2+t^4}{t^2+t^4}$
$\frac{2-3t^2+t^4}{t^2(1+t^2)}=1+\frac{2-4t^2}{t^2+t^4}$
Let $\frac{2-4t^2}{t^2(1+t^2)}=\frac{A}{t^2}+\frac{B}{1+t^2}$
$\Rightarrow A=2,B=\frac{6}{-1}$
$I=-\int (1+\frac{2}{t^2}-\frac{6}{1+t^2})dt$
$\Rightarrow I=-(t-\frac{2}{t}-6{\tan }^{-1}(t\left)\right)+C$
$\therefore I=6{\tan }^{-1}\left(\cos x\right)+2(\cos x{)}^{-1}-\cos x+C$