I=∫_ π/2^π/2 cos x/1+e^xdx=∫_ π/2^π/2cos( x)/1+e^ xdx=∫_π/2^π/2 e^x.cos x/e^x 1dx 2I=∫_ π/2^π/2(1+e^x)cos x/1+e^xdx=2 ∫_ π/2^0 cos x dx=2[sin x]_0^π/2=2 ⇒ I=1 ...

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Standard XIII

Mathematics

Integration of Piecewise Continuous Functions

∫-π/2π/2 cos ...

Question

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{c o s x}{1 + e^{x}} d x =$

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Solution

The correct option is D 1
$I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{c o s x}{1 + e^{x}} d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{c o s (- x)}{1 + e^{- x}} d x = \int_{\frac{π}{2}}^{\frac{π}{2}} \frac{e^{x} . c o s x}{e^{x} - 1} d x$
$2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{(1 + e^{x}) c o s x}{1 + e^{x}} d x = 2 \int_{- \frac{π}{2}}^{0} c o s x d x = 2 [s i n x]_{0}^{\frac{π}{2}} = 2 \Rightarrow I = 1$

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∫π2−π2 cos x1+exdx=

The correct option is D 1I=∫π2−π2 cos x1+exdx=∫π2−π2cos(−x)1+e−xdx=∫π2π2 ex.cos xex−1dx 2I=∫π2−π2(1+ex)cos x1+exdx=2 ∫0−π2 cos x dx=2[sin x]π20=2⇒I=1

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{c o s x}{1 + e^{x}} d x =$