# Evaluate the definite Integral

• Dec 5th 2009, 12:51 AM
zorro
Evaluate the definite Integral
Question : Evaluate

$\displaystyle \int_{-1}^{1} \ \frac{3 + sin^3 x}{1 + x^2} \ dx$
• Dec 5th 2009, 01:08 AM
flyingsquirrel
Quote:

Originally Posted by zorro
Question : Evaluate

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

$\displaystyle \int_{-1}^{1} \frac{3 + \sin^3 x}{1 + x^2} \ \mathrm{d}x=3\int_{-1}^{1} \frac{1}{1 + x^2} \ \mathrm{d}x+\int_{-1}^{1} \frac{\sin^3 x}{1 + x^2} \ \mathrm{d}x$
Hint : note that $\displaystyle x\mapsto \frac{\sin^3 x}{1 + x^2}$ is odd.
• Dec 8th 2009, 12:27 AM
zorro
Stuck again !!!
Quote:

Originally Posted by flyingsquirrel
$\displaystyle \int_{-1}^{1} \frac{3 + \sin^3 x}{1 + x^2} \ \mathrm{d}x=3\int_{-1}^{1} \frac{1}{1 + x^2} \ \mathrm{d}x+\int_{-1}^{1} \frac{\sin^3 x}{1 + x^2} \ \mathrm{d}x$
Hint : note that $\displaystyle x\mapsto \frac{\sin^3 x}{1 + x^2}$ is odd.

I am able to integrate the first term but not the last one

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

= $\displaystyle I_1 + I_2$

$\displaystyle I_1$ = $\displaystyle 3 \left[ tan^{-1} (1) - tan^{-1} (-1) \right] + c_1$

= $\displaystyle 3 \left[ tan^{-1} (1) - tan^{-1} (-1) \right] + c_1$

= $\displaystyle 3 \left[ \frac{\pi}{4} +\frac{\pi}{4} \right] + c_1$

= $\displaystyle 3 \left[ \frac{2 \pi}{4} \right] + c_1$

= $\displaystyle \left[ \frac{6 \pi}{4} \right] + c_1$

$\displaystyle I_2$ = $\displaystyle \int_{-1}^{1} \frac{sin^3 x}{1 + x^2}$ $\displaystyle dx_2$

= $\displaystyle \int_{-1}^{1} \frac{sin^3 x}{1 + x^2}$ $\displaystyle dx_2$

I am stuck here!!!!!......
• Dec 8th 2009, 01:59 AM
dedust
Quote:

Originally Posted by flyingsquirrel
Hint : note that $\displaystyle x\mapsto \frac{\sin^3 x}{1 + x^2}$ is odd.

so $\displaystyle \int_{-1}^{1} \frac{\sin^3 x}{1 + x^2} \ \mathrm{d}x = 0$
• Dec 8th 2009, 02:03 AM
flyingsquirrel
Quote:

Originally Posted by zorro
$\displaystyle I_1$ = $\displaystyle 3 \left[ tan^{-1} (1) - tan^{-1} (-1) \right] + c_1$

There is no need to add a constant since $\displaystyle \int_{-1}^1\frac{1}{1+x^2}\,\mathrm{d}x$ is a definite integral.
Quote:

Originally Posted by zorro
= $\displaystyle \left[ \frac{6 \pi}{4} \right] + c_1$

$\displaystyle \frac{6\pi}{4}=\frac{3\times 2\pi}{2\times 2}=\frac{3\pi}{2}$
Quote:

Originally Posted by zorro
$\displaystyle I_2$ = $\displaystyle \int_{-1}^{1} \frac{sin^3 x}{1 + x^2}$ $\displaystyle dx_2$

$\displaystyle I_2 = \int_{-1}^0\frac{\sin^3x}{1+x^2}\,\mathrm{d}x + \int_0^1\frac{\sin^3x}{1+x^2}\,\mathrm{d}x$
Substitute $\displaystyle x=-t$ ($\displaystyle \implies t=-x$ and $\displaystyle \mathrm{d}t=-\mathrm{d}x$) in the first integral :
$\displaystyle \int_{-1}^0\frac{\sin^3x}{1+x^2}\,\mathrm{d}x =-\int_{1}^0\frac{\sin^3(-t)}{1+(-t)^2}\,\mathrm{d}t=\int_0^1\frac{-\sin^3t}{1+t^2}\,\mathrm{d}t=-\int_0^1\frac{\sin^3t}{1+t^2}\,\mathrm{d}t$
Thus
$\displaystyle I_2 = -\int_0^1\frac{\sin^3t}{1+t^2}\,\mathrm{d}t+\int_0^ 1\frac{\sin^3x}{1+x^2}\,\mathrm{d}x=0.$

This is a special case of a more general result: Let $\displaystyle f:\mathbb{R}\to\mathbb{R}$ be an odd function and let $\displaystyle A$ be a real number. Then
$\displaystyle \int_{-A}^Af(t)\,\mathrm{d}t=0.$

Edit :
Quote:

Originally Posted by dedust
so $\displaystyle \int_{-1}^{1} \frac{\sin^3 x}{1 + x^2} \ \mathrm{d}x = 0$

Yes !
• Dec 8th 2009, 02:17 AM
zorro
Thanks u very much
Thank u all very much for helping me(Bow)
• Dec 8th 2009, 05:48 AM
Krizalid
as for that property about the odd function, we also require that $\displaystyle f$ must be integrable there.
• Dec 8th 2009, 12:01 PM
zorro
what do u mean by that
Quote:

Originally Posted by Krizalid
as for that property about the odd function, we also require that $\displaystyle f$ must be integrable there.

could u please explain it in non mathematical terms ,as i m not that bright??
• Dec 21st 2009, 03:30 PM
zorro
Got it
Quote:

Originally Posted by zorro
could u please explain it in non mathematical terms ,as i m not that bright??

Thanks u all for helping (Beer)