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

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.

• Beyond the problem, take a look at Bioche's rules. These are rules that allow us to reduce ourselves to integrate a rational fraction. May 2 '19 at 20:42

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: 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 $$[-a, a]$$.

• Different to what? May 2 '19 at 20:22
• @LuminousNutria Original integral. May 2 '19 at 20:23

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 $$[-1,1]$$?

• I don't know what a symmetric integral is. May 2 '19 at 20:22
• @LuminousNutria Something of the form $$\int_{-a}^a f(x)\mathrm dx$$ Maybe it would be better to call it an integral with symmetric boundaries. May 2 '19 at 20:22
• @LuminousNutria Yes; more or less ^^' I would call an interval of the form $[-a,a]$ or $[-b,b]$, respectively, symmetric where on the other hand something like $[a,a]$ doesn't really belong to this kind. Do you know how to reduce the integral $$\int_{-a}^af(x)\mathrm dx$$ where $f(x)$ is an odd function, i.e. $f(-x)=-f(x)$? May 2 '19 at 20:27
• @LuminousNutria It's a fundamental property of integrals of odd functions! However, it's quite simple too. See my edit :) And see here for instance. May 2 '19 at 20:36
• @LuminousNutria Glad to help! :) May 2 '19 at 20:45

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.

• If this function is $f(x)$ then are you saying $F(x)-F(-x)=0$? May 2 '19 at 20:42
• Yes, if $f$ is odd. May 2 '19 at 21:00
• Try it with other odd functions, like $\sin(x)$, or $x^3$. May 2 '19 at 21:01