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I am trying to evaluate $\int_{-\pi/4}^{\pi/4}(x^3+x^4\tan x)\,dx.$

I tried to factor out an $x^3$ but that did nothing for me, and I also attempted to split it up into $x^3$ and $x^4 \tan x$ but that didn't help and I am not sure if it is legal. Nothing really seems to be quite right for this.

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    This integral has no simple antiderivative. It does have a particular property that is very useful though. What kind of a function is $x^3+x^4\tan(x)$, on the interval $[-\pi/4,\pi/4]$?2012-04-27

3 Answers 3

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Clearly the integral is zero being an integral of an odd function over a symmetric domain: $ \int_{-\pi/4}^{\pi/4} \left(x^3 + x^4 \tan(x) \right) \mathrm{d} x \stackrel{\color\maroon{x \to -x}}{=} -\int_{-\pi/4}^{\pi/4} \left(x^3 + x^4 \tan(x) \right) \mathrm{d} x $

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    @Jordan: You would know it's an odd function from knowing that $x^3$ is an odd power of $x$, hence an odd function, and $\tan(x)$ is an odd function.2012-04-27
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$\int_{-\pi/4}^{\pi/4} \left(x^3 + x^4 \tan x \right) dx = \int_{-\pi/4}^{\pi/4} x^3 dx + \int_{-\pi/4}^{\pi/4}x^4 \tan x$

If $f(x)=x^3$ then clearly $f(-x)=-f(x)$ (i.e. it is an odd function). Thus, because the absolute value of the bounds are equal

$\int_{-\pi/4}^{\pi/4} x^3 dx = 0$

Letting $g(x)=x^4, h(x)=\tan x$, we have $g(-x)=g(x)$ and $h(-x)=-h(x)$. Thus

$g(-x)h(-x)=-g(x)h(x)$

confirming that $x^4\tan x$ is odd as well, so

$\int_{-\pi/4}^{\pi/4}x^4 \tan x=0$

Confirming that

$\int_{-\pi/4}^{\pi/4} \left(x^3 + x^4 \tan x \right) dx =0$

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The function $f(x) = x^3 + x^4\tan(x)$ i odd. We check this by replacing the $x$ by a $-x$: $f(-x) = (-x)^3 + (-x)^4\tan(-x) = -x^3 + x^4\cdot(-\tan(x)) = -f(x).$ We have used that $\tan$ is an odd function.

Hence you have an integral over an interval symmetric around zero, and so the integral is zero.