Question

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

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

Answer 1

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

$I=\int \frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}dx{\sin }^{8}x-{\cos }^{8}x=({\sin }^{4}x+{\cos }^{4}x)({\sin }^{4}x-{\cos }^{4}x)=(1-2{\sin }^{2}x{\cos }^{2}x)({\sin }^{2}x+{\cos }^{2}x)({\sin }^{2}x-{\cos }^{2}x)=(1-2{\sin }^{2}x{\cos }^{2}x)(-\cos 2x)$
$\therefore I=\int \frac{(1-2{\sin }^{2}x{\cos }^{2}x)(-\cos 2x)}{1-2{\sin }^{2}x{\cos }^{2}x}dx=\int (-\cos 2x)dx=-\frac{1}{2}\sin 2x+C$