Question

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

• 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

Answer: (B)

$I=\int \frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}dx$
$=\int \frac{\left({\sin }^{4}x-{\cos }^{4}x\right)\left({\sin }^{4}x+{\cos }^{4}x\right)}{1-2{\sin }^{2}x{\cos }^{2}x}dx$
$=\int \frac{\left({\sin }^{2}x-{\cos }^{2}x\right)\left({\sin }^{2}x+{\cos }^{2}x\right)\left({\sin }^{4}x+{\cos }^{4}x\right)}{1-2{\sin }^{2}x{\cos }^{2}x}dx$
$1.\left({\sin }^{2}x-{\cos }^{2}x\right)[{\left({\sin }^{2}x+{\cos }^{2}x\right)}^2$
$=\int \frac{-2{\sin }^{2}x{\cos }^{2}x]}{1-2{\sin }^{2}x{\cos }^{2}x}$
$=\int \frac{\left({\sin }^{2}x-{\cos }^{2}x\right)\left(1-2{\sin }^{2}x{\cos }^{2}x\right)}{1-2{\sin }^{2}x{\cos }^{2}x}dx$
$=-\int \cos 2xdx=-\frac{1}{2}\sin 2x+C$