Question

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

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

Answer 1

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

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

$\int \frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx=\int \frac{2\sin x\cos xdx}{1+{\cos }^{2}2x}=-\frac{1}{2}\int \frac{d\left(\cos 2x\right)}{1+(\cos 2x{)}^2}=-\frac{1}{2}{\text{tan}}^{-1}\left(\cos 2x\right)+c$