Question

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

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

Answer 1

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

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

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

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

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

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