The value of $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{{1 + {{\sin }^2}x} \over {1 + {\pi ^{\sin x}}}}} \right)} \,dx$$ is
Solution
$$I = \int\limits_0^{{\pi \over 2}} {{{(1 + {{\sin }^2}x)} \over {(1 + {\pi ^{\sin x}})}} + {{{\pi ^{\sin x}}(1 + {{\sin }^2}x)} \over {(1 + {\pi ^{\sin x}})}}} dx$$<br><br>$I = \int_0^{{\pi \over 2}} {(1 + {{\sin }^2}x)\,dx}$<br><br>$I = {\pi \over 2} + {\pi \over 2}.{1 \over 2} = {{3\pi } \over 4}$
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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