How to solve $\int^1_{-1} \frac{\sin(x)}{1+x^2}dx$?

Question

I have to solve $$\int^1_{-1} \frac{\sin(x)}{1+x^2}\,dx$$

I am a Calculus 1 student, and I am having difficulty because I can't think of anything that I could make into a substitute which would cancel much. I think this may be a difficult problem to solve without using techniques that are beyond a college Calculus 1 level of skill, but please try, or I may have a hard time understanding what you mean.

Here is some of what I've tried:

$u=1+x^2$

$du = 2xdx$

$\frac{du}{2x} = dx$

$$\int^1_{-1}\frac{1}{u} \cdot \sin(x) \cdot \frac{1}{2x} \cdot du$$

I have tried plugging this into Symbolab.com, but it wont even give me a hint what $u$ should be.

Answer 1

HINT

Since $\sin(-x)=-\sin(x)$ you do not need any substitution at all. What do you know about an integral with symmetric boundaries of an odd function?

---

For a given odd function $f(x)$, i.e. $f(-x)=-f(x)$, integrated over a symmetric interval $[-a,a]$, note that we get the following by enforcing the substitution $x\mapsto -x$

\begin{align*} \int_{-a}^af(x)\mathrm dx&=\int_a^{-a}f(-x)(-1)\mathrm dx\\ &=\int_{-a}^af(-x)\mathrm dx\\ &=-\int_{-a}^af(x)\mathrm dx\\ \therefore~2\int_{-a}^af(x)\mathrm dx&=0\\ \end{align*}

$$\therefore~\int_{-a}^af(x)\mathrm dx~=~0$$

Now consider the function $f(x)=\frac{\sin x}{1+x^2}$. Is this one odd; if so what is the integral over the interval $[-1,1]$?

Answer 2

Hint: Try $u = -x$. You may get something very similar but different. Actually, this substitution will give $I = -I$, if $I$ is the original integral.
You can solve that integral when the integration interval is given as $[-a, a]$.

Answer 3

The integrand is an odd function. Odd functions have the property the $f(-x)=-f(x)$. You do not need to find an actual antiderivative to solve this problem, since the integrand on $[-a,0)$ is the negative of the function on $(0,a]$. Therefore, the answer is zero.

Answer 4

Since, $\frac{\sin(-x)}{1+(-x)^2}=-\frac{\sin(x)}{1+x^2}$ the function is symmetric over the interval of $[-1,1]$. Therefore we can evaluate the integral as:

$$\int^1_{-1} \frac{\sin(x)}{1+x^2}\,dx=0$$

Here is a visual representation of what I mean: