Question

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

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

Answer 1

The correct option is C $\sin x-2(\sin x{)}^{-1}-6{\tan }^{-1}(\sin x)+C$

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

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

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

Substitute $\sin x=t$

$\Rightarrow \cos xdx=dt$

The integral becomes $\int \frac{(1-t^2)(2-t^2)}{t^2(1+t^2)}dt$

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

$=\int \frac{t^2(1+t^2)-4(t^2+1)+6}{t^2(1+t^2)}dt$

$=\int (1-\frac{4}{t^2}+\frac{6}{t^2(1+t^2)})dt$

$=\int (1-\frac{4}{t^2}+6(\frac{1}{t^2}-\frac{1}{1+t^2}))dt$

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

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