Question
Integrate the function $\int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx$
=$\int {\frac{{\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^2}x - {{\cos }^2}x} \right)}}{{1 - 2{{\sin }^2}x.{{\cos }^2}x}}} dx$
=$\int {\frac{{\left( {1 - 2{{\sin }^2}x.{{\cos }^2}x} \right)\left( {{{\sin }^2}x - {{\cos }^2}x} \right)}}{{\left( {1 - 2{{\sin }^2}x.{{\cos }^2}x} \right)}}} dx$
=$\int { \left( {{{\sin }^2}x - {{\cos }^2}x} \right)} dx$
=$\int { - \cos 2xdx} $
=$- \frac{{\sin 2x}}{2} + c$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.