Question

Integrals are not always easy to evaluate; sometimes we need to be clever! In this problem we study the definite integral

I=∫(π>0) xsin(x)/1+cos^2(x) dx.

At first, it appears difficult to evaluate this integral with substitution (you can try with simple substitution ideas...). But it can be done! Let's see how it goes.

(a) Use the substitution u=π−x to show that

I= π/2∫(π>0)sin(x)/1+cos^2(x) dx.

(Recall that sin(π−x)=sin(x) and cos(π−x)=−cos(x).)

(b) Evaluate the remaining definite integral using substitution to get:

I=π^2/4.

Answer 1

Let $I = \int_{0}^{π}\frac{xsinx}{1+cos²x}dx$

Let us substitute u = π-x

So du = -dx and x=0 => u = π and x = π => u = 0

$I = \int_{0}^{π}\frac{xsinx}{1+cos²x}dx=-\int_{π}^{0}\frac{(π-u)sin(π-u)}{1+cos²(π-u)} du$ = $- \int_{π}^{0}\frac{(π-u)sinu}{1+cos²u}du$ since sin(π-u)=sinu and cos²(π-u)={-cos(u)}²=cos²u

So $I=\int_{0}^{π}\frac{(π-u)sinu}{1+cos²u}du$ [by property of definite integral that $\int_{a}^{b}f(x)dx = - \int_{b}^{a}f(x)dx$ ]

Therefore $I=\int_{0}^{π}\frac{πsinu}{1+cos²u}du-\int_{0}^{π}\frac{usinu}{1+cos²u}du$

=> $I=π\int_{0}^{π}\frac{sinu}{1+cos²u}du-I$ , since $\int_{0}^{π}\frac{usinu}{1+cos²u}du = \int_{0}^{π}\frac{xsinx}{1+cos²x}dx$

=> $2I= π\int_{0}^{π}\frac{sinu}{1+cos²u}du$

=> $I= \frac{π}{2}\int_{0}^{π}\frac{sinu}{1+cos²u}du$

Let -cos u = z

So sin u du = dz and also when u=0 , z= -1 and when u = π , z = 1

Therefore $I= \frac{π}{2}\int_{-1}^{1}\frac{dz}{1+(-z)²} = \frac{π}{2}\int_{-1}^{1}\frac{dz}{1+z²}$

=> $I = \frac{π}{2}[tan^{-1}z]_{-1}^{1}$

=> $I = \frac{π}{2}[tan^{-1}1 - tan^{-1}(-1)]$

=> $I = \frac{π}{2}(\frac{π}{4}- \frac{-π}{4})$

=> $I = \frac{π}{2}(\frac{π}{4}+ \frac{π}{4} )= \frac{π}{2}.\frac{π}{2} = \frac{π²}{4}$