Question

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

• A. $\frac{1}{2}\sin 2x+c$
• B. $-\frac{1}{2}\sin 2x+c$
• C. $-\frac{1}{2}\sin x+c$
• D. $-si{n}^{2}x+c$

Answer 1

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

$\int \frac{si{n}^{8}x-co{s}^{8}x}{1-2si{n}^{2}xco{s}^{2}x}\cdot dx$

$\int \frac{(si{n}^{4}x+co{s}^{4}x)(si{n}^{2}x-co{s}^{2}x)}{1-2si{n}^{2}x-co{s}^{2}x}dx$

$=-\int \frac{(si{n}^{4}x+co{s}^{4}x)co{s}^{2}x}{si{n}^{4}x+co{s}^{4}x}dx$

$\frac{-1}{2}sin2x+c$