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. $\frac{1}{2}$
• C. $-1$
• D. $1$

Answer 1

The correct option is A $-\frac{1}{2}$

Let $P=\frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}=\frac{({\sin }^{4}x+{\cos }^{4}x)({\sin }^{4}x-{\cos }^{4}x)}{1-2{\sin }^{2}x{\cos }^{2}x}$

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

Therefore,

$I=\int Pdx=\int (-\cos 2x)dx=\frac{-\sin 2x}{2}+C=A+B$

So, $A=\frac{-1}{2}$

Hence, this is the answer.