Definite Integration

**varunnayudu** · December 19th 2008, 04:41 AM

Evaluate
$\int\limits_{-1}^{1}\frac{3+sin^3 x}{1+x^2}dx$

**mr fantastic** · December 19th 2008, 05:05 AM

Originally Posted by varunnayudu

Evaluate
$\int\limits_{-1}^{1}\frac{3 - sin^3 x}{1+x^2}dx$

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

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

**varunnayudu** · January 4th 2009, 08:59 PM

Originally Posted by mr fantastic

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

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

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hw did u get

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

**Danny** · January 4th 2009, 09:35 PM

Originally Posted by varunnayudu

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hw did u get

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

He didn't say

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

He said that

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

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

$\int_{-a}^a f(x) \, dx = 0.$

**Moo** · January 4th 2009, 11:09 PM

If you're not convinced, substitute $t=-x$

Thread: Definite Integration

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