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Let f(x)=(sin^(2)pix)/(1+pi^(x)). Then, ...

Let `f(x)=(sin^(2)pix)/(1+pi^(x)).` Then, `int[f(x)+f(-x)]dx` is equal to

A

0

B

`(1)/(2)x-(sin 2pix)/(4pi)+C`

C

`(x)/(2)-(cospix)/(2pi)+C`

D

`(1)/(1-pi^(x)).(cos^(2)pix)/(2pi)+C`

Text Solution

Verified by Experts

The correct Answer is:
B
(b) `f(x)+f(-x)=(sin^(2)pix)/(1+pi^(x))+(sin^(2)(-pix))/(1+pi^(-x))`
`=(sin^(2)pix)/(1+pi^(x))+((sin^(2)pix)pi^(x))/(pi^(x)+1)`
`=sin^(2)pix=(1-cos 2pix)/(2)`
`therefore int[f(x)+f(-x)]dx=int[(1-cos 2pix)/(2)]dx`
`=(1)/(2)xx-(sin 2pix)/(4pi)+C`
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