JEE MAIN - Mathematics (2018 - 15th April Morning Slot - No. 16)
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 :
$$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 + \log \left( {{{2 + \sin x} \over {2 - \sin x}}} \right)} \right)dx$$ is :
0
$${3 \over 4}$$
$${3 \over 8}$$ $$\pi $$
$${3 \over 16}$$ $$\pi $$
Explanation
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 ]
= $${{3\pi } \over 8}$$
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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