# Thread: Evaluate the definite Integral

1. ## Evaluate the definite Integral

Question : Evaluate

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

2. Originally Posted by zorro
Question : Evaluate

$
\int_{-1}^{1} \ \frac{3 + sin^3 x}{1 + x^2} \ dx
$
$
\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 $x\mapsto \frac{\sin^3 x}{1 + x^2}$ is odd.

3. ## Stuck again !!!

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

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

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

= $I_1 + I_2$

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

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

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

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

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

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

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

I am stuck here!!!!!......

4. Originally Posted by flyingsquirrel
Hint : note that $x\mapsto \frac{\sin^3 x}{1 + x^2}$ is odd.
so $\int_{-1}^{1} \frac{\sin^3 x}{1 + x^2} \ \mathrm{d}x
= 0$

5. Originally Posted by zorro
$I_1$ = $3 \left[ tan^{-1} (1) - tan^{-1} (-1) \right] + c_1$
There is no need to add a constant since $\int_{-1}^1\frac{1}{1+x^2}\,\mathrm{d}x$ is a definite integral.
Originally Posted by zorro
= $\left[ \frac{6 \pi}{4} \right] + c_1$
$\frac{6\pi}{4}=\frac{3\times 2\pi}{2\times 2}=\frac{3\pi}{2}$
Originally Posted by zorro
$I_2$ = $\int_{-1}^{1} \frac{sin^3 x}{1 + x^2}$ $dx_2$
$I_2 = \int_{-1}^0\frac{\sin^3x}{1+x^2}\,\mathrm{d}x + \int_0^1\frac{\sin^3x}{1+x^2}\,\mathrm{d}x$
Substitute $x=-t$ ( $\implies t=-x$ and $\mathrm{d}t=-\mathrm{d}x$) in the first integral :
$\int_{-1}^0\frac{\sin^3x}{1+x^2}\,\mathrm{d}x =-\int_{1}^0\frac{\sin^3(-t)}{1+(-t)^2}\,\mathrm{d}t=\int_0^1\frac{-\sin^3t}{1+t^2}\,\mathrm{d}t=-\int_0^1\frac{\sin^3t}{1+t^2}\,\mathrm{d}t$
Thus
$I_2 = -\int_0^1\frac{\sin^3t}{1+t^2}\,\mathrm{d}t+\int_0^ 1\frac{\sin^3x}{1+x^2}\,\mathrm{d}x=0.$

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

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

6. ## Thanks u very much

Thank u all very much for helping me

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

8. ## what do u mean by that

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

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

9. ## Got it

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

Thanks u all for helping