MHB What is the solution to this week's challenging integral problem?

[anemone]

Here is this week's POTW:

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Compute $$\int_{0}^{\pi} \dfrac{2\sin x+3\cos x-3}{13\cos x-5} \,dx$$.

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[anemone]

Congratulations to the following members for their correct solution::)

1. Opalg
2. greg1313

Solution from Opalg:

Spoiler

At first sight, this looks like an improper integral, because the denominator vanishes when $\cos x = \frac5{13}$. But if $\cos x = \frac5{13}$ then (from the 5-12-13 pythagorean triple) $\sin x = \frac{12}{13},$ which means that $2\sin x + 3\cos x -3 = 0.$ So the numerator also vanishes at that point, and in fact there is no singularity.

To evaluate the integral, use the magical substitution $t = \tan\frac x2$. Then $\sin x = \frac{2t}{1+t^2}$, $\cos x = \frac{1-t^2}{1+t^2}$ and $dx = \frac2{1+t^2}\,dt$. So $$\frac{2\sin x + 3\cos x -3}{13\cos x - 5} = \frac{4t + 3(1-t^2) - 3(1+t^2)}{13(1-t^2) - 5(1+t^2)} = \frac{4t - 6t^2}{8-18t^2} = \frac{2t(2-3t)}{2(2-3t)(2+3t)} = \frac{t}{2+3t},$$ and the integral becomes $$\int_0^\pi \frac{2\sin x + 3\cos x -3}{13\cos x - 5}\,dx = \int_0^\infty\frac t{2+3t}\frac2{1+t^2}\,dt.$$ Next, use partial fractions to get $$\frac{2t}{(2+3t)(1+t^2)} = -\frac{12}{13}\frac1{2+3t} + \frac4{13}\frac t{1+t^2} + \frac6{13}\frac 1{1+t^2}.$$ That function has an indefinite integral $$f(t) = -\frac4{13}\ln(2+3t) + \frac2{13}\ln(1+t^2) + \frac6{13}\arctan t = \frac2{13}\ln\Bigl(\frac{1+t^2}{(2+3t)^2}\Bigr) + \frac6{13}\arctan t.$$

At the lower end of the interval of integration, $$f(0) = -\frac4{13}\ln2$$. At the upper end, $$\lim_{t\to\infty}\frac{1+t^2}{(2+3t)^2} = \frac19,$$ and so $$\lim_{t\to\infty}f(t) = \frac2{13}\ln\Bigl(\frac19\Bigr) + \frac6{13}\frac\pi2 = -\frac4{13}\ln3 +\frac{3\pi}{13}.$$

In conclusion, $$\int_0^\pi \frac{2\sin x + 3\cos x -3}{13\cos x - 5}\,dx = -\frac4{13}\ln3 +\frac{3\pi}{13} + \frac4{13}\ln2 = \frac{3\pi - 4\ln(3/2)}{13}.$$