1. ## Definite Integration

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

2. 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.

3. ## hw did u get this

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.
$

4. 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

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

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