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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 $[-\pi/4,\pi/4]$? – Alex R. Apr 27 '12 at 20:37

3 Answers 3

up vote 10 down vote accepted

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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How would I know it is an odd function before beginning the problem? – user138246 Apr 27 '12 at 20:40
Well, $x^3$ is odd, $\tan(x)$ is odd and $x^4$ is even. It just takes a little bit of practice. You could also get a hint at this by plotting the function. – Sasha Apr 27 '12 at 20:41
@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. – Arturo Magidin Apr 27 '12 at 20:41

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

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


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.

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