Question

The value of the integral $\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\left( {{x^2} + ln\dfrac{{\pi + x}}{{\pi - x}}} \right)cosxdx}$ is :
A) $0$
B) $\dfrac{{{\pi ^2}}}{2} - 4$
C) $\dfrac{{{\pi ^2}}}{2} + 4$
D) $\dfrac{{{\pi ^2}}}{2}$

Answer 1

To evaluate a definite integral the first thing that we’re going to do is evaluate the indefinite integral for the function. This should explain the similarity in the notations for the indefinite and definite integrals.
As you practice definite integrals problems, you will understand that you don’t really need a formula because you know that when doing definite integrals all you need to do is evaluate the indefinite integral and then do the evaluation.

Complete step-by-step answer:
$\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\left( {{x^2} + ln\dfrac{{\pi + x}}{{\pi - x}}} \right)cosxdx}$
Since the derivative of a sum is the sum of the derivatives, the integral of a sum is the sum of the integrals:
$= \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {{x^2}cosxdx} + \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {ln\dfrac{{\pi + x}}{{\pi - x}}\;cosxdx}$
Lets put $f(x) = ln\dfrac{{\pi + x}}{{\pi - x}}\;cosx$
So, $f( - x) = ln\dfrac{{\pi - x}}{{\pi + x}}\;cos( - x)$
$f( - x) = - ln\dfrac{{\pi + x}}{{\pi - x}}\;cos(x)dx = f(x)$
Which implies that f(x) is an odd function. So we can say$\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {ln\dfrac{{\pi + x}}{{\pi - x}}\;cosxdx}$=0
$= \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {{x^2}cosxdx} + 0 = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {{x^2}cosxdx}$
Now, lets put $g(x) = {x^2}cosx$
So, $g( - x) = {\left( { - x} \right)^2}cos\left( { - x} \right)$
$g( - x) = {x^2}\cos x = g(x)$
Which implies that g(x) is an even function.
$= 2\int\limits_{0}^{\dfrac{\pi }{2}} {{x^2}cosxdx}$
Now we will use integration by parts to solve this integration:-
$= 2\left[ {\left( {{x^2}\sin x} \right)_0^{\dfrac{\pi }{2}} - \int\limits_{0}^{\dfrac{\pi }{2}} {2x\sin xdx} } \right]$
Now putting the limits:-
$= 2\left[ {\left( {\dfrac{{{\pi ^2}}}{4}\sin \left( {\dfrac{\pi }{2}} \right) - 0} \right) - 2\int\limits_{0}^{\dfrac{\pi }{2}} {x\sin xdx} } \right]$
Solving the first term and using integration by parts on second term we get
$= \dfrac{{{\pi ^2}}}{2} - 4\left[ {\left( {x( - \cos x)} \right)_0^{\dfrac{\pi }{2}} - \int\limits_{0}^{\dfrac{\pi }{2}} {1( - \cos x)dx} } \right]$
$= \dfrac{{{\pi ^2}}}{2} - 4(0) + 4\int\limits_{0}^{\dfrac{\pi }{2}} {\cos xdx}$
$= \dfrac{{{\pi ^2}}}{2} + 4\left( {\sin x} \right)_0^{\dfrac{\pi }{2}}$
$= \dfrac{{{\pi ^2}}}{2} + 4$

So, option (C) is the correct answer.

Note: When you are faced with an integral the first thing that you need to decide is if there is more than one way to do the integral. If there is more than one way then you need to determine which method to use. The general rule of thumb that you can use is that you should use the method that you find easiest. This may not be the method that others find easiest, but that doesn’t make it the wrong method.