# Evaluate : $\int_0^{\large\frac{\pi}{2}} \large\frac{x+sinx}{1+cosx}$$dx ## 1 Answer Toolbox: • \int udv=uv-\int vdu • \tan\large\frac{\pi}{4}$$=1$
• $\sin x=2\sin \Large\frac{x}{2}$$\cos\Large\frac{x}{2} • 1+\cos x=2\cos^2\Large\frac{x}{2} Step 1: I=\int_0^{\large\frac{\pi}{2}}\large\frac{x+\sin x}{1+\cos x}$$dx$
But $1+\cos x=2\cos^2\Large\frac{x}{2}$
$\sin x=2\sin \Large\frac{x}{2}$$\cos\Large\frac{x}{2} I=\large\frac{1}{2}\int_0^{\Large\frac{\pi}{2}}$$x\sec^2\Large\frac{x}{2}$$dx+\int_0^{\Large\frac{\pi}{2}}\tan\large\frac{x}{2}$$dx$
Let $I=I_1+I_2$
Step 2:
On integrating we get,
Consider $I_1=\int x\sec^2\Large\frac{x}{2}$$dx Let u=x\Rightarrow du=dx \sec^2\Large\frac{x}{2}$$dx=dv$
$2\tan \Large\frac{x}{2}$$=v \int udv=uv-\int vdu On substituting for u,v,du and dv we get, (x)(2\tan\Large\frac{x}{2})$$-\int 2\tan \Large\frac{x}{2}$$dx \therefore I=\large\frac{1}{2}$$\big[2x\tan\Large\frac{x}{2}\big]_0^{\Large\frac{\pi}{2}}-\frac{1}{2}$$\times 2\int_0^{\Large\frac{\pi}{2}}\tan x/2dx+\int_0^{\Large\frac{\pi}{2}}\tan x/2dx \big[x\tan\Large\frac{x}{2}\big]_0^{\Large\frac{\pi}{2}}-\int_0^{\Large\frac{\pi}{2}}$$\tan x/2dx+\int_0^{\Large\frac{\pi}{2}}\tan x/2dx$
Step 3:
On applying limits
$\large\frac{\pi}{2}$$\tan\Large\frac{\pi}{4}$$-0$