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Integral question I have to integrate this function $$I = \int_0^4\frac{20x-5x^2}{x^2+9} \mathrm{d}x$$ obtaining $$20\ln(5/3) + 15\tan^{-1}(4/3) -20.$$ However, my calculator, even after somehow simplifying it a bit, gives this: $$\frac{40\ln(5/3) -30\tan^{-1}(3/4) +15\pi -40}{2}$$

As you can see there is something wrong with arctan integration, can anybody help or know how to simplify this with some identity?

I am asking this because it is one of the questions in the practise exam, and in the exam i will have to use this calculator, no others allowed, so if a question like this comes up, i will get stuck..

Eric Towers
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Deo
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2 Answers2

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The answers are the same when you notice that, for $x\gt0$, $$ \tan^{-1}\left(\frac1x\right)=\frac\pi2-\tan^{-1}(x) $$

Note that $$ \tan^{-1}\left(\frac ba\right)=B\quad\text{and}\quad\tan^{-1}\left(\frac ab\right)=A $$ and $$ A+B=\frac\pi2 $$ $\hspace{3.4cm}$enter image description here

If you want to use trigonometric identities, $$ \begin{align} \tan\left(\frac\pi2-x\right) &=\frac{\sin\left(\frac\pi2-x\right)}{\cos\left(\frac\pi2-x\right)}\\ &=\frac{\sin\left(\frac\pi2\right)\cos(x)-\cos\left(\frac\pi2\right)\sin(x)}{\cos\left(\frac\pi2\right)\cos(x)+\sin\left(\frac\pi2\right)\sin(x)}\\[6pt] &=\frac{\cos(x)}{\sin(x)}\\[12pt] &=\frac1{\tan(x)} \end{align} $$

robjohn
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  • @robjohn: Caught my typo a second after adding the comment. Deleted subsequently. – Eric Towers Apr 26 '14 at 22:46
  • Thanks a lot for this identity. But I am still a bit confused of how it is true... – Deo Apr 26 '14 at 22:51
  • @Deo: consider a right triangle. The non-right angles add up to $\frac\pi2$. If you compute the tangents of the non-right angles you are dividing the non-hypontenuse sides in opposite orders. – robjohn Apr 26 '14 at 22:53
  • Ok so tan(x) = 3/4 tan(pi/2 - x) = 4/3 is that it? Is there a trigonometric identity way to prove it? – Deo Apr 26 '14 at 22:57
  • @Deo: I've added a diagram – robjohn Apr 26 '14 at 23:03
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By first simplifying the fraction, we get that $$\frac{40\ln(5/3) -30\tan^{-1}(3/4) +15\pi -40}{2} = 20\ln(5/3) -15\tan^{-1}(3/4) +\frac{15}{2}\pi -20$$

Subtracting $20\ln(5/3) + 15\tan^{-1}(4/3) -20$ from it, we get $$(20\ln(5/3) -15\tan^{-1}(3/4) +\frac{15}{2}\pi -20) - (20\ln(5/3) + 15\tan^{-1}(4/3) -20)$$ $$ = \frac{15}{2}\pi - 15\tan^{-1}(4/3) -15\tan^{-1}(3/4) = 15 \left(\frac{\pi}{2} - \tan^{-1} (4/3) - \tan^{-1} (3/4)\right)$$

Here we bring in the definition of tangent. Given a right triangle $ABC$ with $C$ being the right angle, $$\tan \angle ABC = \frac{AC}{BC}$$ From this, we get that $$\angle ABC = \tan^{-1} \frac{AC}{BC}$$

Now, we take a $3-4-5$ triangle and note that the two inverse tangents are the two angles that are not $\displaystyle \frac{\pi}{2}$. Because the angles of a triangle add up to $\pi$, we note that they sum to $\displaystyle \frac{\pi}{2}$. Therefore, $$15 \left(\frac{\pi}{2} - \tan^{-1} (4/3) - \tan^{-1} (3/4)\right) = 15 \left(\frac{\pi}{2} - \frac{\pi}{2}\right) = 0$$

Q.E.D.

2012ssohn
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