Question

Integration of sin^8x-cos^8x/1-2sin^2xcos^2x

Answer 1

$\underline{\textsf{Solution :}}$

$\mathsf{Now,\:\int \frac{sin^{8}x-cos^{8}x}{1-2sin^{2}xcos^{2}x}dx}$

$\mathsf{=\int \frac{(sin^{4}x+cos^{4}x)(sin^{4}x-cos^{4}x)}{1-2sin^{2}xcos^{2}x}dx}$

$\tiny{\mathsf{=\int \frac{\{(sin^{2}x+cos^{2}x)^{2}-2sin^{2}xcos^{2}x\}(sin^{2}x+cos^{2}x)(sin^{2}x-cos^{2}x)}{1-2sin^{2}xcos^{2}x}dx}}$

$\mathsf{=\int \frac{(1-2sin^{2}xcos^{2}x)(sin^{2}x-cos^{2}x)}{1-2sin^{2}xcos^{2}x}dx}$

$\mathsf{=\int (sin^{2}x-cos^{2}x)dx}$

$\mathsf{=-\int (cos^{2}x-sin^{2}x)dx}$

$\mathsf{=-\int cos2x\:dx}$

$\mathsf{=-\frac{sin2x}{2}+C}$

$\textsf{where C is integral constant.}$

$\to \boxed{\small{\mathsf{\int \frac{sin^{8}x-cos^{8}x}{1-2sin^{2}xcos^{2}x}dx = -\frac{sin2x}{2}+C}}}$

$\textsf{Thus, solved.}$