How to solve this integral $\int \frac 1{\sqrt { \cos x \sin^3 x }} \mathrm dx $

Question

Question : $$\int \frac 1{\sqrt { \cos x \sin^3 x }} \mathrm dx $$

I don’t know where to start. I had tried many methods but they didn’t work.

Can anyone help me solving this ? Thank you

Answer 1

HINT: substitute $\text{u}:=\tan\left(x\right)$. Then the integrand will change to $\frac{1}{\text{u}^\frac{3}{2}}$.

Answer 2

Hint :

1) Double angle formulas bring the expression to be integrated under the form :

$$\dfrac{2}{\sqrt{\sin(2x)(1-\cos(2x))}}$$

2) Then use formulas :

$$\cos(a)=\dfrac{1-t^2}{1+t^2} \ \ \ \sin(a)=\dfrac{2t}{1+t^2} \ \ \ \text{with} \ a=2x$$

where $t=\tan(a/2)=\tan(x)$ (thus with $x=\arctan(t)$ whence $dx=\dfrac{dt}{1+t^2}$).

Answer 3

$\int \frac 1{\sqrt { \cos x \sin^3 x }} \mathrm dx$

multiply by $\sec^{2}x$ then you get

$\int \frac{\sec^{2}x}{\sqrt{\tan^{3}x}} \mathrm dx$

now let $\tan x=t$

One gets $\int\frac{dt}{\sqrt{t^{3}}}$

Now I think u can do it