A) \[\frac{3}{16}\pi \]
B) 0
C) \[\frac{3}{8}\pi \]
D) \[\frac{3}{4}\]
Correct Answer: A
Solution :
Let \[I=\int_{-\frac{\pi }{2}}^{\frac{x}{2}}{{{\sin }^{4}}x\left( 1+\log \left( \frac{2+\sin x}{2-\sin x} \right) \right)dx}\]------\[i\] Now we know that, \[\int_{a}^{b}{f(x).dx=\int_{a}^{b}{f(a+b-x).dx}}\] \[\Rightarrow I=\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{({{\sin }^{4}}(-x)).(1+\log (\frac{1+\sin (-x)}{1-\sin (-x)})).dx}\] \[=\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{({{\sin }^{4}}x).(1+\log (\frac{1-\sin x}{1+\sin x})).dx}\] \[=\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{{{\sin }^{4}}x.(1-\log (\frac{1+\sin x}{1-\sin x})).dx}\]-------\[ii\] Adding equation i and ii we get, \[2I=\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{{{\sin }^{4}}x.dx}\] \[2I=2\int_{0}^{\frac{x}{2}}{{{\sin }^{4}}x.dx}\] \[I=\int_{0}^{\frac{\pi }{2}}{{{\sin }^{4}}x.dx}\] Use the reduction formula \[\int_{{}}^{{}}{{{\sin }^{m}}x.dx=\frac{-\cos x.si{{n}^{m-1}}(x)}{m}+\frac{m-1}{m}\int_{{}}^{{}}{{{\sin }^{-2+m}}x.dx}}\] Here, \[\begin{align} & \int_{0}^{\frac{\pi }{2}}{{{\sin }^{4}}x.dx=\frac{-\cos x.{{\sin }^{3}}x}{4}|_{0}^{\frac{\pi }{2}}+\frac{3}{4}\int_{0}^{\frac{\pi }{2}}{{{\sin }^{2}}x.dx}} \\ & ----As\frac{-\cos x.{{\sin }^{3}}x}{4}|_{0}^{\frac{\pi }{2}}=0 \\ \end{align}\]\[\Rightarrow \frac{3}{4}\int_{0}^{\frac{\pi }{2}}{\frac{1-\cos 2x}{2}.dx}\] \[=\frac{3}{8}\int_{0}^{\frac{\pi }{2}}{1.dx-\frac{3}{8}\int_{0}^{\frac{x}{2}}{\cos 2x.dx}}\] After putting the limits we get, \[=\frac{3\pi }{16}-----As(\int_{0}^{\frac{\pi }{2}}{\cos 2x.dx)=0}\] \[\Rightarrow \frac{3\pi }{16}\]
You need to login to perform this action.
You will be redirected in
3 sec
