Question

If $\int \frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}dx=A\sin 2x+B$, then $A=$

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

Answer 1

Answer: (A)

$-\frac{1}{2}$

${\sin }^{8}x-{\cos }^{8}x=({\sin }^{4}x+{\cos }^{4}x)({\sin }^{4}x-{\cos }^{4}x)$

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

$=(1-2{\sin }^{2}x{\cos }^{2}x)(-\cos 2x)$

$\therefore \int \frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}dx=-\int \cos 2xdx=-\frac{1}{2}\sin 2x+B$, where $B$ is constant