May be a tough integral

Evaluate $\int \frac{d\theta}{\cos^3 \theta + \sin^3 \theta} $

Answer:

Substitute $\sin \theta = \frac{2t}{t^2 + 1} $, thus we have

$\sin^2 \theta = \frac{4t^2}{(t^2 + 1)^2} $

and

$\cos^2 \theta = 1 - \frac{4t^2}{(t^2 + 1)^2} = \frac{(t^2 - 1)^2}{(t^2 + 1)^2}$

or, $\cos \theta = \frac{t^2 - 1}{t^2 + 1} $

Also, from $\sin \theta = \frac{2t}{t^2 + 1} $, we have

$\cos \theta \; d\theta = \frac{2(t^2+1) - 4t^2}{(t^2 + 1)^2} \; dt = \frac{2(1 - t^2)}{(t^2 + 1)^2} \; dt$

or, $d \theta = \frac{-2}{t^2 + 1} \; dt $

So, $\int \frac{d\theta}{\cos^3 \theta + \sin^3 \theta} $

$= \int \frac{1}{\frac{8t^3}{(t^2 + 1)^3} + \frac{(t^2 - 1)^3}{(t^2 + 1)^3}} \;\frac{-2}{t^2 + 1} \; dt $

$= \int \frac{-2(t^2 + 1)^2}{8t^3 + (t^2 - 1)^3} \; dt $

We should be able to integrate it using partial fractions method.

So, I used Wolfram and little bit of my own simplification to get

$ \frac{(t^2 + 1)^2}{8t^3 + (t^2 - 1)^3} $

$= \frac{1+\sqrt{3}}{6} \frac{1} {\left(t - \frac{\sqrt{3} + 1}{2}\right)^2 + \frac{2 + \sqrt{3}}{2}} $

$+ \frac{1-\sqrt{3}}{6} \frac{1} {\left(t + \frac{\sqrt{3} - 1}{2}\right)^2 + \frac{2 - \sqrt{3}}{2}} $

$+ \frac{1}{3\sqrt{2}} \frac{1}{t + \sqrt{2} + 1} $

$+ \frac{1}{3\sqrt{2}} \frac{1}{t - \sqrt{2} + 1} $

So, $ \int \frac{-2(t^2 + 1)^2}{8t^3 + (t^2 - 1)^3} \; dt$

$=\frac{-1-\sqrt{3}}{3} \frac{\sqrt{2}}{\sqrt{2 + \sqrt{3}}} \tan^{-1} \frac{t - \frac{\sqrt{3} + 1}{2}} { \sqrt{ \frac{2 + \sqrt{3}} {2} }} $

$+\frac{\sqrt{3} - 1}{3} \frac{\sqrt{2}}{\sqrt{2 - \sqrt{3}}} \tan^{-1} \frac{t + \frac{\sqrt{3} - 1}{2}} { \sqrt{ \frac{2 - \sqrt{3}} {2} }} $

$-\frac{\sqrt{2}}{3} \ln (t + \sqrt{2} + 1) $

$-\frac{\sqrt{2}}{3} \ln (t - \sqrt{2} + 1) + C$

Since $t = \cot \frac{\theta}{2} $, we substitute back to get

$\int \frac{d\theta}{\cos^3 \theta + \sin^3 \theta} $

$=\frac{-1-\sqrt{3}}{3} \frac{\sqrt{2}}{\sqrt{2 + \sqrt{3}}} \tan^{-1} \frac{\cot \frac{\theta}{2} - \frac{\sqrt{3} + 1}{2}} { \sqrt{ \frac{2 + \sqrt{3}} {2} }} $

$+\frac{\sqrt{3} - 1}{3} \frac{\sqrt{2}}{\sqrt{2 - \sqrt{3}}} \tan^{-1} \frac{\cot \frac{\theta}{2} + \frac{\sqrt{3} - 1}{2}} { \sqrt{ \frac{2 - \sqrt{3}} {2} }} $

$-\frac{\sqrt{2}}{3} \ln (\cot \frac{\theta}{2} + \sqrt{2} + 1) $

$-\frac{\sqrt{2}}{3} \ln (\cot \frac{\theta}{2} - \sqrt{2} + 1) + C$