Integrating the given function involving trigonometric functions

Question

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.

Answer 1

$$ \int \csc ^{p/3}(x) \sec^{q/3}(x)dx = \int \frac{1}{\sin^{p/3}(x)\cos^{4-\frac{p}{3}}(x)}dx = \int \frac{1}{\cos^4 x \cdot \tan^{p/3}{x}}dx = \int \frac{\sec^4 x}{\tan ^{p/3}x}dx $$

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

Answer 2

HINT:

$$\csc^{p/3}x\sec^{q/3}=\dfrac1{\sin^{p/3}x\cos^{(12-p)/3}x}$$

$$=\dfrac{\sec^2x}{\tan^{p/3}x\cos^2x}=\dfrac{(1+\tan^2x)\sec^2x}{\tan^{p/3}x}$$

Set $\tan x=u$

In general, $$\dfrac1{\sin^px\cos^{2I-p}x}=\cdots=\dfrac{(1+\tan^2x)^{I-1}\sec^2x}{\tan^px}$$