Integrals — Class 12 Maths Solution

: Let$I = \int {\cfrac{{\cos 2x}}{{{{\left( {\cos x + \sin x} \right)}^2}}}dx = \int {\cfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\left( {\cos x + \sin x} \right)}^2}}}} } dx$

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

Put ${\mathop{\rm cosx}\nolimits} + sinx = t$ $\Rightarrow$ $\left( { - \sin x + \cos x} \right)dx = dt$

$\therefore$ $I = \int {\cfrac{{dt}}{t}} = \log \left| t \right| + C = \log \left| {\cos x + \sin x} \right| + C$