Integrating the given function involving trigonometric functions Find $\int \csc^{p/3}x \sec^{q/3} x dx $ 
Given - $(p,q \in I^{+} )$ and $(p+q=12)$
I tried to substitute $q = 12-p$ in the integral but didn't find anything satisfactory.

$$
\int \csc ^{p/3}(x) \sec^{q/3}(x)dx = \int \frac{1}{\sin^{p/3}(x)\cos^{4-\frac{p}{3}}(x)}dx = \int \frac{1}{\cos^4 x \cdot \tan^{p/3}{x}}dx = \int \frac{\sec^4 x}{\tan ^{p/3}x}dx 
$$
Let $u=\tan x \implies du = \sec^2 x dx \implies$
$$
\int\frac{(1+u^2)du}{u^{p/3}} = \frac{3}{3-p}u^{\frac{3-p}{3}} + \frac{3}{9-p}u^{\frac{9-p}{3}} + C = \frac{3}{3-p}\tan^{\frac{3-p}{3}}x + \frac{3}{9-p}\tan^{\frac{9-p}{3}}x + C 
$$
