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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.

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

enter image description here

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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 $[-a, a]$.

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  • $\begingroup$ Different to what? $\endgroup$ May 2 '19 at 20:22
  • $\begingroup$ @LuminousNutria Original integral. $\endgroup$
    – Seewoo Lee
    May 2 '19 at 20:23
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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 $[-1,1]$?

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  • $\begingroup$ I don't know what a symmetric integral is. $\endgroup$ May 2 '19 at 20:22
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    $\begingroup$ @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. $\endgroup$
    – mrtaurho
    May 2 '19 at 20:22
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    $\begingroup$ @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. $f(-x)=-f(x)$? $\endgroup$
    – mrtaurho
    May 2 '19 at 20:27
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    $\begingroup$ @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. $\endgroup$
    – mrtaurho
    May 2 '19 at 20:36
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    $\begingroup$ @LuminousNutria Glad to help! :) $\endgroup$
    – mrtaurho
    May 2 '19 at 20:45
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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.

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

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