# Definite Integration

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• Dec 19th 2008, 04:41 AM
varunnayudu
Definite Integration
Evaluate
$\displaystyle \int\limits_{-1}^{1}\frac{3+sin^3 x}{1+x^2}dx$
• Dec 19th 2008, 05:05 AM
mr fantastic
Quote:

Originally Posted by varunnayudu
Evaluate
$\displaystyle \int\limits_{-1}^{1}\frac{3 - sin^3 x}{1+x^2}dx$

Note that $\displaystyle \int\limits_{-1}^{1}\frac{\sin^3 x}{1+x^2} \, dx = 0$ since $\displaystyle \sin^3 x$ is an odd function.

So $\displaystyle \int\limits_{-1}^{1}\frac{3 - \sin^3 x}{1+x^2} \, dx = \int\limits_{-1}^{1}\frac{3}{1+x^2} \, dx$ which you should be able to easily do.
• Jan 4th 2009, 08:59 PM
varunnayudu
hw did u get this
Quote:

Originally Posted by mr fantastic
Note that $\displaystyle \int\limits_{-1}^{1}\frac{\sin^3 x}{1+x^2} \, dx = 0$ since $\displaystyle \sin^3 x$ is an odd function.

So $\displaystyle \int\limits_{-1}^{1}\frac{3 - \sin^3 x}{1+x^2} \, dx = \int\limits_{-1}^{1}\frac{3}{1+x^2} \, dx$ which you should be able to easily do.

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hw did u get

$\displaystyle \frac{sin^3x}{1+x} = 0.$
• Jan 4th 2009, 09:35 PM
Jester
Quote:

Originally Posted by varunnayudu
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hw did u get

$\displaystyle \frac{sin^3x}{1+x} = 0.$

He didn't say

$\displaystyle \frac{sin^3x}{1+x^2} = 0.$ (note the number 2)

He said that

$\displaystyle \int_{-1}^1 \frac{sin^3x}{1+x^2}\, dx = 0.$

In fact, if $\displaystyle f(x)$ is odd and continuous on $\displaystyle [-a,a]$ then

$\displaystyle \int_{-a}^a f(x) \, dx = 0.$
• Jan 4th 2009, 11:09 PM
Moo
If you're not convinced, substitute $\displaystyle t=-x$ :D