Thread: Evaluate the definite Integral

Question : Evaluate

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

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.

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!!!!!......

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 $

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.

Originally Posted by zorro

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

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

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 :

Originally Posted by dedust

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

Yes !

Thank u all very much for helping me

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

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??

Originally Posted by zorro

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

Thanks u all for helping