Question

Evaluate $\int \frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$

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

Answer 1

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

Given,

$\int \frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}dx$

Divide numerator and denominater by $co{s}^{4}x$

The integral becomes

$\therefore \int \frac{tanx.se{c}^{2}x}{1+ta{n}^{4}x}dx$

Let

$ta{n}^{2}x=t⟹2tanxSe{c}^{2}xdx=dt⟹tanxse{c}^{2}x=\frac{dt}{2}$

$\therefore \int \frac{1}{2}\frac{1}{1+t^2}dt=\frac{1}{2}ta{n}^{-1}t+c=\frac{1}{2}ta{n}^{-1}\left(ta{n}^{2}x\right)+c$ ( as $t=ta{n}^{2}x$)