# Integration — can't figure out the substitution

I need to evaluate the integral $$\int_0^1 \frac{6\pi}{\lvert{2 - 3e^{2\pi i t}}\rvert^2} \,\mathrm{d}t.$$ The problem I am having is that I can't find a nice way to re-write the denominator to invoke a substitution. Perhaps I can view this as a contour integral in the complex plane? Any ideas appreciated.

• Contour integral won't likely work nice as the integrand is a real function... – DonAntonio Apr 4 '17 at 18:07
• @DonAntonio by contour I mean techniques like integrating $\int_\infty^\infty \sin x/x dx$ using $e^z/z$ – user369210 Apr 5 '17 at 1:42
• Well, you can see in my answer that you can now substitute $\;z=e^{2\pi it}\;$ and integrate around the unit circle in the complex plane... – DonAntonio Apr 5 '17 at 8:37


• Too clever; is this a technique that you find widely applicable? – user369210 Apr 9 '17 at 4:17
• @user321210 Thanks. When Residue Theorem isn't 'available' it's better to try another approach. This one yields a straightforward answer. – Felix Marin Apr 9 '17 at 22:20

An idea:

$$2-3e^{2\pi it}=2-3\cos2\pi t-3i\sin2\pi t\implies |2-3e^{2\pi it}|^2=13-12\cos2\pi t\implies$$$${}$$

$$\int_0^1\frac{6\pi}{|2-3e^{2\pi it}|^2}dt=\int_0^1\frac{6\pi}{13-12\cos2\pi t}dt$$

You can now try some trigonometric substitution.

• And now contour integration is viable. ;-)) (+1) – Mark Viola Apr 4 '17 at 18:10
• @Dr.MV Shhh.... :) . Indeed so, and it will probably be the easiest way. – DonAntonio Apr 4 '17 at 18:17