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I have been trying to solve the following integral for a little while now

$$I = \int_a^b \arccos\left( \frac{x}{\sqrt{(a+b)x - ab \,}\,}\right)\mathrm{d}x \quad 0<a<b$$

After doing some testing in maple, I was able to discover that $\displaystyle I=\frac{\pi}{4}\frac{(a-b)^2}{b+a} $

But I want to show this algebraically. (using mathematics)

I have done some effort and one can notice that the denominator equals $x^2-(x-a)(x-b)$. Another thing i tried using was the fact that

$$\arccos(x) = \arcsin\left(\sqrt{1-x^2}\,\right)$$

also if on tries solving it by parts

$$ \left[ \arccos\left( \frac{x}{\sqrt{(a+b)x - ab \,}\,}\right) \right]_a^b = 0 $$

but the remaining integral looks impossible. If one uses the substitution, $u=\text{denominator}$ I end up with

$$ \frac{2}{a+b}\int u^2 \arccos\left( \frac{u^2 + ab}{u(b+a)} \right) \, \mathrm{d}u$$

Also $$I = \int_a^b \text{arccsc}\left( \sqrt{\frac{(a-x)(b-x)}{x^2}}\right)\mathrm{d}x$$

not sure what this buys me though. I also tried a variety of clever things such at differentiating under the integral sign, but alas nothing worked. Anyone mind helping me ? =)

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3 Answers 3

up vote 11 down vote accepted

The answer is $$\int_{x=a}^b \arccos\left({x\over\sqrt{(a+b)x-ab}}\right)\,dx={\pi (b-a)^2 \over 4(a+b)}.$$

This is problem 11457 from the American Mathematical Monthly. The problem was published in the October 2009 issue, and the solution appeared in the April 2011 issue.

Using integration by parts $\int u \,dv=uv-\int v\,du$ with $u=\arccos\left({x\over\sqrt{(a+b)x-ab}}\right)$ and $v=x-ab/(a+b)$ we have $$\int_{a}^b \arccos\left({x\over\sqrt{(a+b)x-ab}}\right)\,dx =\int_a^b {(b+a)x-2ab\over 2(a+b)\sqrt{(b-x)(x-a)}}\,dx .\tag1 $$ Using the change of variables $w=(2x-(a+b))/(b-a)$, the integral on the right hand side of (1) becomes $$\int_{-1}^1{(b^2-a^2)w+(b-a)^2 \over 4(a+b)\sqrt{1-w^2}}\,dw ={ (b-a)^2 \over 4(a+b) } \int_{-1}^1{dw \over \sqrt{1-w^2}} ={\pi (b-a)^2 \over 4(a+b)},$$ since the integral of the odd function $w/\sqrt{1-w^2}$ over the interval $(-1,1)$ is zero, and $\int_{-1}^1 (1-w^2)^{-1/2}\,dw=\arcsin(1)-\arcsin(-1)=\pi$.


enter image description here

The integrand is the angle $\theta(x)$ in this diagram. I tried to find a geometric explanation for the OP's integral, but I failed. Maybe someone else can succeed?

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Very nice, is there any reason for choosing $w = (2x-(a+b))/(b-a)$ other than it obviously works? =) –  N3buchadnezzar Mar 19 '12 at 16:29
    
@N3buchadnezzar I fooled around with a lot of different substitutions before I hit on this one. No special insight: just trial and error! –  Byron Schmuland Mar 19 '12 at 16:50

A succinct and beautiful solution has already posted while I was writing this answer, so I'm not sure if it will be helpful...

Let $\alpha = \sqrt{b/a} > 1$ and $\beta = \frac{1}{2}(\alpha - \alpha^{-1})$. Then by the substitution $ x = \sqrt{ab} u$, it suffices to show that

$$\int_{\alpha^{-1}}^{\alpha} \arccos \left( \frac{u}{\sqrt{(\alpha + \alpha^{-1}) u - 1}} \right) \; du = \frac{\pi \beta^2}{\alpha + \alpha^{-1}}.$$

Now by the substitution $v = \sqrt{(\alpha + \alpha^{-1}) u - 1}$, it is equivalent to

$$\int_{\alpha^{-1}}^{\alpha} 2 v \arccos \left( \frac{v + v^{-1}}{\alpha + \alpha^{-1}} \right) \; dv = \pi \beta^2 .$$

Now performing integration by parts, the left hand side becomes

$$\int_{\alpha^{-1}}^{\alpha} \frac{v^2 - 1}{\sqrt{(\alpha + \alpha^{-1})^2 - (v + v^{-1})^2}} \; dv = \int_{\alpha^{-1}}^{\alpha} \frac{v^2 - 1}{\sqrt{(\alpha - \alpha^{-1})^2 - (v - v^{-1})^2}} \; dv .$$

Finally, plug $v = w + \sqrt{1+w^2}$. Then this integral reduces to

$$\int_{-\beta}^{\beta} \frac{w\left(w+\sqrt{1+w^2}\right)^2}{\sqrt{\beta^2 - w^2}\sqrt{1+w^2}} \; dw .$$

Now expanding the square in the numerator, the odd terms cancel out, yielding

$$\int_{-\beta}^{\beta} \frac{2w^2}{\sqrt{\beta^2 - w^2}} \; dw .$$

This integral is easily calculated by the substitution $w = \beta \sin\theta$, verifying the identity as desired.

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+1 This is very nice! –  Byron Schmuland Mar 19 '12 at 17:42

Did you try that? As my first intuition:

$\int \arccos\left(x\,/\sqrt{(a+b)x-ab\,}\,\right)\,\mathrm {d}x=x.\arccos\left(x\,/\sqrt{(a+b)x-ab\,}\,\right)+\int x.\frac{\left(x\,/\sqrt{(a+b)x-ab\,}\,\right)'}{\sqrt{1-\frac{x^2}{(a+b)x-ab}}}\,\mathrm {d}x$

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