The value of ∫−π2π211+esinxdx is

I=∫−π2π211+esinxdx ∫abf(x)dx=∫abf(a+b−x)dx I=∫−π2π211+esinx I=∫−π2π2esinxesinx+1dx 2I=∫−π2π2dx 2I=2π2 I=π2

The value of ∫−π2π211+esinxdx is

I=∫−π2π211+esinxdx ∫abf(x)dx=∫abf(a+b−x)dx I=∫−π2π211+esinx I=∫−π2π2esinxesinx+1dx 2I=∫−π2π2dx 2I=2π2 I=π2

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The value of $\int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{1}{1 + e^{\sin x}} d x$ is

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$π$

$\frac{π}{4}$

$\frac{3 π}{2}$

$\frac{π}{2}$

answer is D.

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Detailed Solution

$I = \int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{1}{1 + e^{\sin x}} d x \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x I = \int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{1}{1 + e^{\sin x}} I = \int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{e^{\sin x}}{e^{\sin x} + 1} d x 2 I = \int_{\frac{- π}{2}}^{\frac{π}{2}} d x 2 I = \frac{2 π}{2} I = \frac{π}{2}$