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In my homework I need to solve the integral: (Homework, so I would like a hint more than a full solution)

$$ \int_{-\pi/4}^{\pi/4}\frac{x^7- x + 1}{\cos^2x}~\mathrm dx $$

What I tried:

We can use integration by parts, knowing that:

$$ (\tan x)' = \frac{1}{\cos^2x} $$

Therefore it seems fit to take:

$$ u = (x^7 - x + 1), v' = \frac{1}{cos^2x} $$

Now doing the integration I get:

$$ \int_{-\pi/4}^{\pi/4}\frac{x^7- x + 1}{\cos^2x}~\mathrm dx = (x^7-x+1)\tan^2x - \int_{-\pi/4}^{\pi/4}(7x^6 - 1)\tan^2x~\mathrm dx $$

Now I need to solve the new right term I get:

$$ \int_{-\pi/4}^{\pi/4}(7x^6 -1)\tan^2x~\mathrm dx $$

And again, it seems that it goes the way of integration by parts, but, because of the $x^6$ I will need again and again to do the integration by parts...

So, whats wrong?

Simply Beautiful Art
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Alon
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3 Answers3

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$$ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}{\frac{x^{7}-x+1}{\cos^{2}{x}}\,\mathrm{d}x}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}{\frac{x^{7}-x}{\cos^{2}{x}}\,\mathrm{d}x}+\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}{\frac{\mathrm{d}x}{\cos^{2}{x}}} $$

Since $ x\mapsto\frac{x^{7}-x}{\cos^{2}{x}} $ is an odd function, we have that : $$ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}{\frac{x^{7}-x}{\cos^{2}{x}}\,\mathrm{d}x}=0 $$

And $$ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}{\frac{\mathrm{d}x}{\cos^{2}{x}}}=\bigg[\tan{x}\bigg]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}=2 $$

Thus $$ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}{\frac{x^{7}-x+1}{\cos^{2}{x}}\,\mathrm{d}x}=2 $$

Simply Beautiful Art
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CHAMSI
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1

$$\dots=\int_{-\pi/4}^{\pi/4}\frac{x^7- x}{\cos^2x}~\mathrm dx+ \int_{-\pi/4}^{\pi/4}\frac{ 1}{\cos^2x}~\mathrm dx=$$

$$ 0+ \int_{-\pi/4}^{\pi/4}\frac{ 1}{\cos^2x}~\mathrm dx= 2\int_{0}^{\pi/4}\frac{ 1}{~\mathrm cos^2x}~\mathrm dx=\dots$$

Simply Beautiful Art
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User2020201
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0

Following your way you get

$$\int_{-\pi/4}^{\pi/4}\frac{x^7- x + 1}{\cos^2x}~\mathrm dx$$ $$ = \left[(x^7-x+1)\color{blue}{\tan x}\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} - \int_{-\pi/4}^{\pi/4}\underbrace{(7x^6 - 1)\color{blue}{\tan x}}_{odd}~\mathrm dx$$ $$= 2-0 = 2$$

trancelocation
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  • Ok thank you very much. Can i ask a basic question in that subject? why the integral is 0? i know there is a negative area and positive that cancel each other, but its "area under the curve" so why are we cancel each other and not taking the || of each, getting double value? – Alon Apr 01 '20 at 13:53
  • If you interpret such an integral as "area under the curve", you need to be aware that areas below the $x$-axis get a negative sign. – trancelocation Apr 01 '20 at 13:55
  • And in general, for a Riemann-integrable odd function $f(-x)=-f(x)$ on $[-a,a]$, you have $\int_{-a}^af(x);dx=0$ – trancelocation Apr 01 '20 at 13:57
  • Ok so if i just interapt that as an integral they cancel, and if as are under the curve they sum as i thought, wierd notion, so what is an integral if not area under the curve?... i hate not understanding what im doing,,, thanks again – Alon Apr 01 '20 at 13:59
  • It IS area under the curve as long as the function is ABOVE the $x$-axis. Just take for example $f(x) = x$ on $[-1,1]$ and see what happens. If you reall want only the areas (without sign), you would integrate $|x|$ to assure that you really get only the area. Otherwise, you would get $0$. – trancelocation Apr 01 '20 at 14:02