Integrals — Class 12 Maths Solution

Let $I = \int {\cfrac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx$

We have, $\left( {{{\sin }^8}x - {{\cos }^8}x} \right) = \left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( {{{\sin }^4}x - {{\cos }^4}x} \right)$

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

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

$\therefore$ $I = \int {\cfrac{{\left( {1 - 2{{\sin }^2}x{{\cos }^2}x} \right)\left( { - \cos 2x} \right)}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}dx}$

$= - \int {\cos 2x} dx = - \cfrac{1}{2}\sin 2x + C$