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$$\int_{-1}^{1} \frac{x^3\,e^{\sin \left( \cos \left( x \right) \right)}+1}{x^2+1}\mathrm{d}x$$

$\cos x$ is an even function, $\sin(even)$ is a compostion of of an odd function and an even function which is an even function.

$e^{x}$ is neither even nor odd, so the function $e^{\sin( \cos ( x ))}$ is even, now $x^3\,e^{\sin \left( \cos \left( x \right) \right)}$ is a product of an even and odd function so it is odd.

Overall we got $\int_{-1}^{1} \frac{odd+1}{even+1}$

What Can we say about $\int_{-1}^{1} \frac{odd+1}{even+1}=\int_{-1}^{1} \frac{odd}{even}$?

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  • $\begingroup$ Can I say that $\frac{odd}{even}=(odd)*(even)^{-1}=(odd)$? $\endgroup$ – gbox Aug 3 '16 at 11:18
  • $\begingroup$ The odd-even argument won't work for this integral. $\endgroup$ – StubbornAtom Aug 3 '16 at 11:25
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$I=\displaystyle\int_{-1}^{1} \frac{x^3\,e^{\sin \left( \cos \left( x \right) \right)}+1}{x^2+1}\,dx$

$=\displaystyle\int_0^1\left (\frac{x^3\,e^{\sin \left( \cos \left( x \right)) \right)}+1}{x^2+1}+\frac{-x^3\,e^{\sin \left( \cos \left( x \right) \right)}+1}{x^2+1}\right )\,dx$ $\quad (*)$

$=\displaystyle\int_0^1 \frac{2}{x^2+1}\,dx=\frac{\pi}{2}$


$(*)\to \quad \int_{-a}^{a}f(x)\,dx=\int_0^a (f(x)+f(-x))\,dx$

(Simply write $\int_{-a}^{a}f=\int_{-a}^0f+\int_0^af$ and then substitute $x=-t$ in the 1st integral.)

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  • $\begingroup$ Why can I do this? the function is odd? $\endgroup$ – gbox Aug 3 '16 at 10:56
  • $\begingroup$ @gbox We use a property of definite integral, nothing more. $\endgroup$ – StubbornAtom Aug 3 '16 at 10:58
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$I=\int_{-1}^1 f(x) dx=\int_{-1}^1 f(-x) dx$ holds (substitution $x\to -x$).

Therefore, $$2I=\int_{-1}^1 (f(x)+f(-x)) dx=\int_{-1}^1 \frac{2}{x^2+1} dx=2\arctan{x}|_{-1}^1,$$ $$I=\arctan{x}|_{-1}^1=\frac{\pi}{2}.$$

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  • $\begingroup$ $f(x)\neq f(-x)$ $\endgroup$ – StubbornAtom Aug 3 '16 at 11:07
  • $\begingroup$ @StubbornAtom I didn't wrote that $f(x)=f(-x)$. As you can see from my answer $f(x)+f(-x)=\frac{2}{x^2+1}$. I wrote for integrals: $\int_{-1}^1 f(x) dx=\int_{-1}^1 f(-x) dx$ which holds, since for substitution $x=-t$, we have: $ \int_{-1}^1 f(x) dx= \int_{1}^{-1} f(-t) d(-t)=\int_{-1}^{1} f(-t) dt=\int_{-1}^{1} f(-x) dx$. $\endgroup$ – alans Aug 3 '16 at 11:12
  • $\begingroup$ Yes you're right. My bad. $\endgroup$ – StubbornAtom Aug 3 '16 at 11:22
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Odd + 1 isn't odd. $\operatorname{id} (x)=x$ is odd, but $x+1$ isn't.

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  • $\begingroup$ Where I wrote odd and even, I mean an odd function not a number $\endgroup$ – gbox Aug 3 '16 at 10:53
  • $\begingroup$ $x+1$ isn't an odd function, while $id(x)=x$ is. So 1 + odd FUNC $\neq$ odd $\endgroup$ – Lan Aug 3 '16 at 11:08
  • $\begingroup$ And a composition of ANY function with an even function is an even function. $\endgroup$ – Lan Aug 3 '16 at 11:15
  • $\begingroup$ odd/even=odd always holds: let $f=odd/even=o/e$ then $f(-x)=o(-x)/e(-x)=-o(x)/e(x)=-f(x)$ so f is odd $\endgroup$ – Lan Aug 3 '16 at 11:29

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