Mathematics

# $\displaystyle \int _{ -\pi /2 }^{ \pi /2 }{ \dfrac { \cos { x } }{ 1+{ e }^{ x } } } dx$

##### SOLUTION
$\displaystyle I=\, ^{\frac{\pi }{2}} _{\frac{-\pi }{2}} \int \frac{\cos x}{1+e^{x}}dx$      ......(i)

$\displaystyle \therefore I=\, _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\int \frac{\cos (\frac{\pi }{2}-\frac{\pi }{2}-x)}{1+e(\frac{\pi }{2}-\frac{\pi }{2}-x)}dx$

$\displaystyle=\, _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\int \frac{\cos x}{1+ e^{-x}}dx=\, _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\int \frac{e^{x}\cos x}{1+e^{x}}dx$    ....(ii)

Adding equation (i) and (ii), we get
$\displaystyle \Rightarrow 2I=\, _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\int \cos x dx = [\sin x]_{-\frac{\pi }{2}}^{\frac{\pi }{2}}$
$= \displaystyle \sin\frac{\pi }{2}- \sin \left(-\frac{\pi }{2}\right)$
$1+1$
$=2$
$\Rightarrow 2I=2$
$\Rightarrow I=1$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

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• B. $\log(\sqrt{2}-2)$
• C. $\log(\sqrt{2}-1)$
• D. $\log(\sqrt{2}+1)$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Solve :- $\int\sin^3 x dx$

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Q3 Multiple Correct Hard
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1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
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