Question
The value of the integral $$\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left( {{x^2} + ln\frac{{\pi + x}}{{\pi - x}}} \right)} \cos x\,dx$$ is-
A.
$$0$$
B.
$$\frac{{{\pi ^2}}}{2} - 4$$
C.
$$\frac{{{\pi ^2}}}{2} + 4$$
D.
$$\frac{{{\pi ^2}}}{2}$$
Answer :
$$\frac{{{\pi ^2}}}{2} - 4$$
Solution :
$$\eqalign{
& \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left[ {{x^2} + ln\left( {\frac{{\pi + x}}{{\pi - x}}} \right)} \right]} \cos x\,dx \cr
& = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{x^2}} \cos x\,dx + \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {ln\left( {\frac{{\pi + x}}{{\pi - x}}} \right)} \cos x\,dx \cr
& = 2\int\limits_0^{\frac{\pi }{2}} {{x^2}} \cos \,x\,dx + 0 \cr} $$
[as $${{x^{2\,}}\,\cos \,x}$$ is an even function and $$ln\left( {\frac{{\pi + x}}{{\pi - x}}} \right)\,\cos \,x$$ is an odd function]
$$\eqalign{
& = 2\left[ {{x^2}\sin \,x + 2x\,\cos \,x - 2\,\sin x} \right]_0^{\frac{\pi }{2}} \cr
& = 2\left( {\frac{{{\pi ^2}}}{4} - 2} \right) \cr
& = \frac{{{\pi ^2}}}{2} - 4 \cr} $$