(a)OR(b)

Find: $\displaystyle \int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x$

\[\begin{aligned} \int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x & =\int \frac{\cos ^{2} x-\sin ^{2} x}{(\sin x+\cos x)^{2}} d x \\ & =\int \frac{\cos x-\sin x}{\sin x+\cos x} d x \\ & =\log |\sin x+\cos x|+C \end{aligned} \] \[\begin{aligned} & I=\int_{0}^{\pi / 2} \frac{5 \sin x+3 \cos x}{\sin x+\cos x} d x \\ & I=\int_{0}^{\pi / 2} \frac{5 \sin \left(\dfrac{\pi}{2}-x\right)+3 \cos \left(\dfrac{\pi}{2}-x\right)}{\sin \left(\dfrac{\pi}{2}-x\right)+\cos \left(\dfrac{\pi}{2}-x\right)} d x=\int_{0}^{\pi / 2} \frac{5 \cos x+3 \sin x}{\cos x+\sin x} d x \\ & \text { Adding (i) and (ii), we get } \\ & \left.2 I=\int_{0}^{\pi / 2} 8 d x \Rightarrow I=4 x\right]_{0}^{\frac{\pi}{2}}=2 \pi \end{aligned} \]