Integration of the following trignometry

Question

We have to find the integration of the following ,

I tried but got stuck , can anyone help me

Answer 1

Use Weierstrass substitution

By setting $t= \tan \left( \frac x2 \right) $

The integral is equivalent to:

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

Decompose the fraction into:

$$ 2 \int \left( \frac 1{3-t^2} - \frac 8{(3-t^2)^2} + \frac{16}{(3-t^2)^3} \right)dt$$

Since $\frac 1{3-t^2} = \frac 1{2\sqrt 3} \left( \frac 1{\sqrt 3 - t} + \frac 1{\sqrt 3 +t} \right)$ (Or you can use inverse hyperbolic tangent, which is the same function) Now try substituting $t=\sqrt 3 \sin (v)$ for the other two fractions:

$$\int \frac 1{(3- 3\sin^2 v)^2} \sqrt 3 \operatorname{cos} v dv= \frac{\sqrt{3}}{9}\int \sec^3 v dv$$

The third fraction requires integrating $\sec^5 x$, these two can be solved by integration by parts.

The remaining work is tedious, but not hard. Hope this answer helps.