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Calculus Level 3

$\int _{ -\pi }^{ \pi }{ \frac { { \left( \cos { x } \right) }^{ 2 } }{ 1+{ a }^{ x } } } \ \mathrm dx$

For a constant $a > 0$ , evaluate the integral above.

a\pi

\frac { \pi }{ 2 }

2\pi

\frac { \pi }{ a }

2 solutions

Aniket Verma
Mar 25, 2015

$I = \displaystyle \int _{ -\pi }^{ \pi }{ \frac { { \left( \cos { x } \right) }^{ 2 } }{ 1+{ a }^{ x } } } dx$

$\implies I = \displaystyle\int _{ -\pi }^{ \pi }{ \frac { { \left( \cos { (-x) } \right) }^{ 2 } }{ 1+{ a }^{ -x } } } dx$

$\therefore ~ 2I=\displaystyle\int _{ -\pi }^{ \pi }{ { cos }^{ 2 }x } dx~ =~ 2\int _{ 0 }^{ \pi }{ { cos }^{ 2 }x } dx$

$\implies I=\displaystyle\int _{ 0 }^{ \pi }{ \frac { 1+cos2x }{ 2 } } dx\quad =\quad \left[ \frac { x }{ 2 } +\frac { sin2x }{ 4 } \right] _{ 0 }^{ \pi }=~ \frac { \pi }{ 2 }$

Can yoi explain more

Mohamed Kasim - 6 years, 2 months ago

Just to add to Aniket's comment, note that

$\dfrac{1}{1 + a^{x}} + \dfrac{1}{1 + a^{-x}} = \dfrac{(1 + a^{-x}) + (1 + a^{x})}{(1 + a^{x})(1 + a^{-x})} = \dfrac{2 + a^{x} + a^{-x}}{2 + a^{x} + a^{-x}} = 1.$

Brian Charlesworth - 6 years, 2 months ago

Or, $\displaystyle\dfrac{1}{1+a^x} + \dfrac{1}{1+a^{-x}} = \dfrac{1}{1+a^x} + \dfrac{a^x}{1+a^x} = \dfrac{1+a^x}{1+a^x}=1$

Purushottam Abhisheikh - 6 years, 2 months ago

$\because \displaystyle\int^{b}_{a}~f(x)dx = \int^{b}_{a}~f(b+a-x)dx$ . Using this you will get the second step. After this add first and second step to get the third one. I hope this will help you.

Aniket Verma - 6 years, 2 months ago

Thank you very much

Mohamed Kasim - 6 years, 2 months ago

Shashank Rustagi
Mar 23, 2015

\int _{ -\pi }^{ \pi }{ \frac { { \left( \cos { x } \right) }^{ 2 } }{ 1+{ a }^{ x } } } dx can be written as \int _{ 0 }^{ \pi }{ cos(x) } dx

$\int _{ -\pi }^{ \pi }{ \frac { { \left( \cos { x } \right) }^{ 2 } }{ 1+{ a }^{ x } } } dx$

$\int _{ 0 }^{ \pi }{ cos(x) } dx$

Your Latex code gives the above output . You sure that it's correct ?

A Former Brilliant Member - 6 years, 2 months ago

yes it is correct

Shashank Rustagi - 6 years, 2 months ago

It should be $\displaystyle \int _{ 0 }^{ \pi }{ cos^{2}(x) } dx$ and not $\displaystyle \int _{ 0 }^{ \pi }{ cos(x) } dx$

Purushottam Abhisheikh - 6 years, 2 months ago

Can You solve this? You are great then. Feel Proud

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