JEE MAIN - Mathematics (2021 - 26th August Evening Shift - No. 14)
The value of $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{{1 + {{\sin }^2}x} \over {1 + {\pi ^{\sin x}}}}} \right)} \,dx$$ is
$${\pi \over 2}$$
$${{5\pi } \over 4}$$
$${{3\pi } \over 4}$$
$${{3\pi } \over 2}$$
Explanation
$$I = \int\limits_0^{{\pi \over 2}} {{{(1 + {{\sin }^2}x)} \over {(1 + {\pi ^{\sin x}})}} + {{{\pi ^{\sin x}}(1 + {{\sin }^2}x)} \over {(1 + {\pi ^{\sin x}})}}} dx$$
$$I = \int_0^{{\pi \over 2}} {(1 + {{\sin }^2}x)\,dx} $$
$$I = {\pi \over 2} + {\pi \over 2}.{1 \over 2} = {{3\pi } \over 4}$$
$$I = \int_0^{{\pi \over 2}} {(1 + {{\sin }^2}x)\,dx} $$
$$I = {\pi \over 2} + {\pi \over 2}.{1 \over 2} = {{3\pi } \over 4}$$
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