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. ${\sin }^{2}x+C$

Answer 1

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

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

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

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

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

$=\int \cos 2xdx$

$=\frac{\sin 2x}{2}+C$