Question

Integrate the following functions w.r.t. x.

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

Answer 1

$\int \frac{si{n}^{8}x-co{s}^{8}x}{1-2si{n}^{2}xco{s}^{2}x}dx=\int \frac{\left\{\right(si{n}^{4}x{)}^2-\left(co{s}^{4}x{)}^2\right\}}{1-2si{n}^{2}xco{s}^{2}x}dx=\int \frac{(si{n}^{4}x-co{s}^{4}x)(si{n}^{4}x+co{s}^{4}x)}{1-2si{n}^{2}xco{s}^{2}x}dx[\because a^2-b^2=(a-b\left)\right(a+b\left)\right]=\int \frac{(si{n}^{2}x-co{s}^{2}x)(si{n}^{2}x+co{s}^{2}x)\left[\right(si{n}^{2}x{)}^2+\left(co{s}^{2}x{)}^2\right]}{1-2si{n}^{2}xco{s}^{2}x}dx=\int \frac{(si{n}^{2}x-co{s}^{2}x)1\left\{\right(si{n}^{2}x+co{s}^{2}x{)}^2-2si{n}^{2}x.co{s}^{2}x\}}{1-2si{n}^{2}xco{s}^{2}x}dx[si{n}^{2}x+co{s}^{2}x=1and{a}^2+b^2=(a+b{)}^2-2ab]=\int \frac{-cos2x\{1-2si{n}^{2}x.co{s}^{2}x\}}{1-2si{n}^{2}xco{s}^{2}x}dx(\because cos2x=co{s}^{2}x-si{n}^{2}x)=-\int cos2xdx=-\frac{sin2x}{2}+C$