Mathematics

The value of $\int_{-\dfrac{\pi}{2}} ^ {\dfrac{\pi}{2}} \dfrac{x^2\cos x}{1+e^x}dx$ is equal to

SOLUTION
Let $I=\int _{ \cfrac { -\pi }{ 2 } }^{ \cfrac { \pi }{ 2 } }{ \cfrac { { x }^{ 2 }\cos { x } }{ 1+{ e }^{ x } } dx }$
Using $\int _{ -a }^{ a }{ f(x)dx } =\int _{ 0 }^{ a }{ (f(x)+f(-x))dx }$ , we get
$I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cfrac { { x }^{ 2 }\cos { x } (1+{ e }^{ x })dx }{ (1+{ e }^{ x }) } } =-\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ { x }^{ 2 }\cos { x } dx }$
using product rule, we get
$I={ x }^{ 2 }\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cos { x } dx } -\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ [(\cfrac { d{ x }^{ 2 } }{ dx } )\int { \cos { x } dx } ]dx } \\ I={ x }^{ 2 }{ [\sin { x } ] }_{ 0 }^{ \cfrac { \pi }{ 2 } }-\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ 2x\sin { x } dx } \\ I=\cfrac { { \pi }^{ 2 } }{ 4 } -2[x\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \sin { x } dx } -\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ [\cfrac { dx }{ dx } \int { \sin { x } dx } ]dx } ]\\ I=\cfrac { { \pi }^{ 2 } }{ 4 } -2[0+\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cos { x } dx } ]\\ I=\cfrac { { \pi }^{ 2 } }{ 4 } -2(1)=\cfrac { { \pi }^{ 2 } }{ 4 } -2$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

Realted Questions

Q1 Single Correct Hard
If $x$ satisfies the equation $\displaystyle\left(\int_0^1{\frac{dt}{t^2+2t\cos{\alpha}+1}}\right)x^2-\left(\int_{-3}^3{\frac{t^2\sin{2t}}{t^2+1}dt}\right)x-2=0$
for $(0<\alpha<\pi)$
then the value of $x$ is?
• A. $\displaystyle\pm\sqrt{\frac{\alpha}{2\sin{\alpha}}}$
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• C. $\displaystyle\pm\sqrt{\frac{\alpha}{\sin{\alpha}}}$
• D. $\displaystyle\pm2\sqrt{\frac{\sin{\alpha}}{\alpha}}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \int_{0}^{1}\tan ^{-1}x\:dx.$
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1 Verified Answer | Published on 17th 09, 2020

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