Question

$\int \frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}dx$

• A. $-\frac{1}{2}\sin 2x+C$
• B. $\frac{1}{2}\sin 2x+C$
• C. $\frac{1}{2}\sin x+C$
• D. $-\frac{1}{2}\sin x+C$

Answer 1

The correct option is A $-\frac{1}{2}\sin 2x+C$

Let $I=\int \frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}dx$
As ${\sin }^{8}x-{\cos }^{8}x=({\sin }^{4}x+{\cos }^{4}x)({\sin }^{2}x+{\cos }^{2}x)({\sin }^{2}x-{\cos }^{2}x)$
$=\int \frac{({\sin }^{2}x-{\cos }^{2}x)({\sin }^{4}x+{\cos }^{4}x)}{1-2{\sin }^{2}x{\cos }^{2}x}dx$
Since we can use
$[{({\sin }^{2}x+{\cos }^{2}x)}^2-2{\sin }^{2}x{\cos }^{2}x]=({\sin }^{4}x+{\cos }^{4}x)\left(1\right)$
$=\int \frac{{\sin }^{2}x-{\cos }^{2}x[({\sin }^{2}x+{\cos }^{2}x)-2{\sin }^{2}x{\cos }^{2}x]}{1-2{\sin }^{2}x-{\cos }^{2}x}dx=\int ({\sin }^{2}x-{\cos }^{2}x)dx$
$=-\int \cos 2xdx$
$=-\frac{\sin 2x}{2}+C$