QJEEM18Mab011: JEE Main 2018 Mathematics - Class 12 JEEM18Mab011 Solution: The value of the integral is :

Question

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The value of the integral

\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 + \log \left( {{{2 + \sin x} \over {2 - \sin x}}} \right)} \right)dx

is :

Accepted Answer

Let

I =

\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 + \log \left( {{{2 + \sin x} \over {2 - \sin x}}} \right)} \right)dx

......(1)

\Rightarrow

I =

\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}}(- x)\left( {1 + \log \left( {{{2 + \sin (- x)} \over {2 - \sin(- x)}}} \right)} \right)dx

[ As

\int\limits_a^b { f \left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)} dx

]

=

\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 + \log \left( {{{2 - \sin x} \over {2 + \sin x}}} \right)} \right)dx

=

\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 - \log \left( {{{2 + \sin x} \over {2 - \sin x}}} \right)} \right)dx

.......(2)

Adding equation (1) and (2) we get,

2I = 2

\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} xdx

2I = 4

\int\limits_0^{{\pi \over 2}} {{{\sin }^4}} xdx

I = 2

\int\limits_0^{{\pi \over 2}} {{{\sin }^4}} xdx

=

{{2 \times {3 \over 2} \times {1 \over 2} \times \pi } \over {2 \times 2}}

[ Using Gamma function ]

=

{{3\pi } \over 8}

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