Question

Evaluate the following expression $\int\limits_{0}^{\pi }{\dfrac{x\sin \left( x \right)}{1+{{\cos }^{2}}\left( x \right)}dx}$.

Answer 1

We will us the fact that the $\int\limits_{a}^{b}{f\left( x \right)}dx=\int\limits_{a}^{b}{f\left( b+a-x \right)}dx$ in this problem. We use this because $\sin \left( x \right)$ is an odd function and ${{\cos }^{2}}\left( x \right)$ is an even function.

Complete step-by-step solution
Let $I=\int\limits_{0}^{\pi }{\dfrac{x\sin \left( x \right)}{1+{{\cos }^{2}}\left( x \right)}dx}$. Using the fact that $\int\limits_{a}^{b}{f\left( x \right)}dx=\int\limits_{a}^{b}{f\left( b+a-x \right)}dx$, we can rewrite the integral I as,
$I=\int\limits_{0}^{\pi }{\dfrac{\left( \pi -x \right)\left( \sin \left( \pi -x \right) \right)}{1+{{\cos }^{2}}\left( \pi -x \right)}dx}$
Observe that $\sin \left( \pi -x \right)=\sin \left( x \right)$ and ${{\cos }^{2}}\left( \pi -x \right)={{\left( -\cos \left( x \right) \right)}^{2}}={{\cos }^{2}}\left( x \right)$.
Thus, on substituting this value in the integral, we get that,
$I=\int\limits_{0}^{\pi }{\dfrac{\left( \pi -x \right)\sin \left( x \right)}{1+{{\cos }^{2}}\left( x \right)}dx}$
Separating the terms, we get that the value of the integral is,
$I=\pi \int\limits_{0}^{\pi }{\dfrac{\sin \left( x \right)}{1+{{\cos }^{2}}\left( x \right)}dx}-\int\limits_{0}^{\pi }{\dfrac{x\sin \left( x \right)}{1+{{\cos }^{2}}\left( x \right)}dx}$
But the second term in the right hand side is the required integral I. So on sending it to the left - hand side, we will get that,
$2I=\pi \int\limits_{0}^{\pi }{\dfrac{\sin \left( x \right)}{1+{{\cos }^{2}}\left( x \right)}dx}$
Dividing both the sides by 2, we will get,
$I=\dfrac{\pi }{2}\int\limits_{0}^{\pi }{\dfrac{\sin \left( x \right)}{1+{{\cos }^{2}}\left( x \right)}dx}$
Hence now we just need to evaluate,
${{I}_{1}}=\int\limits_{0}^{\pi }{\dfrac{\sin \left( x \right)}{1+{{\cos }^{2}}\left( x \right)}dx}$
We will solve this integral using substitution.
Let $u=\cos \left( x \right)$. Differentiating both sides, we get that,
$-\sin \left( x \right)dx=du\Rightarrow \sin \left( x \right)dx=-du$
The new limits of the integration will be from $\cos \left( 0 \right)$ to $\cos \left( \pi \right)$ which gives the limits of the integration as 1 and −1.Thus we get that,
${{I}_{1}}=\int\limits_{1}^{-1}{\dfrac{-du}{1+{{u}^{2}}}}$
Changing the limits of the integral, we get that,
${{I}_{1}}=\int\limits_{-1}^{1}{\dfrac{du}{1+{{u}^{2}}}}$
This is a standard integral and we know that,
$\int{\dfrac{dx}{1+{{x}^{2}}}={{\tan }^{-1}}\left( x \right)}$
Using this, we get that,
${{I}_{1}}=\left[ {{\tan }^{-1}}\left( u \right) \right]_{-1}^{1}=\dfrac{\pi }{2}$
The required value of the integral I was $\dfrac{\pi }{2}{{I}_{1}}$. Hence, on substituting the value of ${{I}_{1}}$, we get that,
$I=\int\limits_{0}^{\pi }{\dfrac{x\sin \left( x \right)}{1+{{\cos }^{2}}\left( x \right)}dx}=\dfrac{{{\pi }^{2}}}{4}$

Note: In this problem it is important to observe that x sin(x) is an odd function and ${{\cos }^{2}}x$ is an even function. This is important because once you use this fact, you can break down the given integral into a simpler integral which is much easier to solve. Also while solving the definite integrals, it is important to make sure that you change the values of the limits of the integral otherwise, you might get the value of the integral wrong. Students sometimes miss out on the negative sign that one gets while substituting u = cos(x) and also the one you get while interchanging the limits 1 and -1 of the integral, ${{I}_{1}}$.