What is the Integral of Sin(x)/(x^2+1) from -1 to 1?

[karush]

$\text{S6.7.r.12}$

$$\displaystyle
\int_{-1}^{1}\frac{\sin\left({x}\right)}{1+x^2} \,dx =0 $$

graphing this, it shows an odd function with the values canceling each other out resulting in zero.
but step wise not sure how to take the integral of this.

View attachment 5816

 

[Greg]

I'm pretty sure the integral is non-elementary. Wolfram Alpha gives a nasty result.

 

[karush]

yeah I don't know its an even problem # so no bk answer the TI said it was zero.

online Integral Calculator gave this
$$\dfrac{\mathrm{i}\cosh\left(1\right)\left(\operatorname{Si}\left(x+\mathrm{i}\right)
-\operatorname{Si}\left(x-\mathrm{i}\right)\right)+\sinh\left(1\right)\left(\operatorname{Ci}\left(x+\mathrm{i}\right)+\operatorname{Ci}\left(x-\mathrm{i}\right)\right)}{2}$$

I was clueless

 

[Prove It]

$\text{S6.7.r.12}$

$$\displaystyle
\int_{-1}^{1}\frac{\sin\left({x}\right)}{1+x^2} \,dx =0 $$

graphing this, it shows an odd function with the values canceling each other out resulting in zero.
but step wise not sure how to take the integral of this.

Theorem: If your function f(x) is an odd function (i.e. f(x) = f(-x) for all x) then $\displaystyle \begin{align*} \int_{-a}^a{f(x)\,\mathrm{d}x} = 0 \end{align*}$.

So all you need to do is show that $\displaystyle \begin{align*} \frac{\sin{ \left( -x \right) }}{1 + \left( -x \right) ^2} = -\left[ \frac{\sin{(x)}}{1 + x^2} \right] \end{align*}$...

 

[MarkFL]

Suppose we have an even function $F$:

$$F(x)=F(-x)$$

and we differentiate w.r.t $x$:

$$\frac{d}{dx}F(x)=\frac{d}{dx}F(-x)$$

$$f(x)=-f(-x)$$

We see then that $f$ (the derivative of $F$) must be an odd function, therefore we know the anti-derivative of an odd function is an even function, and thus we may state:

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

This is the odd-function rule for integration.

 

[karush]

I guess the hint was -1 to 1