We are asked to solve the following integral: $\int \frac{\cos x}{\sin^3 x (1-\sin^2 x)^{5/2}} dx$

1. Problem Description

We are asked to solve the following integral:

$\int \frac{\cos x}{\sin^3 x (1-\sin^2 x)^{5/2}} dx$

2. Solution Steps

First, we simplify the integral using the identity $\cos^2 x + \sin^2 x = 1$, which implies $1 - \sin^2 x = \cos^2 x$.

Then the integral becomes:

$\int \frac{\cos x}{\sin^3 x (\cos^2 x)^{5/2}} dx = \int \frac{\cos x}{\sin^3 x \cos^5 x} dx = \int \frac{1}{\sin^3 x \cos^4 x} dx$

Now, we can rewrite the integral as follows:

$\int \frac{1}{\sin^3 x \cos^4 x} dx = \int \frac{1}{\sin^3 x \cos^3 x \cos x} dx = \int \frac{1}{\sin^3 x \cos^3 x} \frac{1}{\cos x} dx = \int \frac{\sin^3 x + \cos^3 x}{\sin^3 x \cos^3 x} \frac{1}{\cos x} dx$

Instead, let's try a substitution. Let $u = \sin x$. Then $du = \cos x dx$. The integral becomes

$\int \frac{du}{u^3 (1-u^2)^{5/2}}$. This seems complicated.

Let's go back to the original substitution idea. Let $u = \sin x$, then $du = \cos x dx$.

Our integral is $\int \frac{\cos x}{\sin^3 x (\cos^2 x)^{5/2}} dx = \int \frac{\cos x}{\sin^3 x \cos^5 x} dx = \int \frac{1}{\sin^3 x \cos^4 x} dx$.

Then the integral can be written as $\int \frac{\cos x}{\sin^3 x (\cos^2 x)^{5/2}}dx = \int \frac{1}{\sin^3 x \cos^4 x} dx = \int \frac{1}{\sin^3 x (1-\sin^2 x)^2} \cos x dx$.

Now we can substitute $u=\sin x$, $du = \cos x dx$.

The integral becomes $\int \frac{1}{u^3 (1-u^2)^2} du = \int \frac{1}{u^3 (1 - 2u^2 + u^4)} du = \int \frac{1}{u^3 - 2u^5 + u^7} du$.

This integral looks difficult to solve directly.

Instead, rewrite the integral as

$\int \frac{1}{\sin^3 x \cos^4 x} dx = \int \frac{\sin^2 x + \cos^2 x}{\sin^3 x \cos^4 x} dx = \int \frac{\sin^2 x}{\sin^3 x \cos^4 x} dx + \int \frac{\cos^2 x}{\sin^3 x \cos^4 x} dx = \int \frac{1}{\sin x \cos^4 x} dx + \int \frac{1}{\sin^3 x \cos^2 x} dx = \int \frac{1}{\sin x \cos^4 x} dx + \int \csc^3 x \sec^2 x dx$.

This also does not seem promising.

Let us rewrite the integral as

$\int \frac{1}{\sin^3 x \cos^4 x} dx = \int \frac{\sin^4 x + \cos^4 x + 2\sin^2 x \cos^2 x}{(\sin x \cos x)^3 \cos x} dx$

Multiply top and bottom by $\cos x$

$\int \frac{1}{\sin^3 x \cos^4 x} dx = \int \frac{\cos x}{\sin^3 x \cos^5 x}dx$

$\int \frac{\cos x}{\sin^3 x (1 - \sin^2 x)^{5/2}} dx$

Let $u = \sin x$, $du = \cos x dx$

$\int \frac{1}{u^3 (1 - u^2)^{5/2}} du$

Let $u = \sin x$.

$I = \int \frac{1}{\sin^3 x \cos^4 x} dx$. $I = \int \frac{1}{\sin^3 x \cos^4 x} dx = \int \frac{1}{\sin^3 x (1 - \sin^2 x)^2} \cos x dx$

$I = \int \frac{1}{u^3 (1 - u^2)^2} du = \int \frac{1}{u^3 (1 - 2u^2 + u^4)} du = \int \frac{1}{u^3 - 2u^5 + u^7} du$

Let $u = \cos x$, so $du = -\sin x dx$.

Rewrite as $\int \frac{\cos x}{\sin^3 x (\cos^2 x)^{5/2}}dx = \int \frac{\cos x dx}{\sin^3 x \cos^5 x} = \int \frac{dx}{\sin^3 x \cos^4 x}$

$\int \frac{\sec^7 x}{(\tan x)^3} dx$

Let $t = \tan x$. Then $dt = \sec^2 x dx$. So $\sec^2 x = 1 + t^2$.

So $\int \frac{\sec^5 x}{(t)^3} dt = \int \frac{(1+t^2)^{5/2}}{t^3} dt$.

This is still hard!

Back to: $\int \frac{1}{u^3 (1 - u^2)^2} du$ where $u = \sin x$

Use partial fractions.

$\frac{1}{u^3 (1-u^2)^2} = \frac{1}{u^3 (1-u)^2 (1+u)^2} = \frac{A}{u} + \frac{B}{u^2} + \frac{C}{u^3} + \frac{D}{1-u} + \frac{E}{(1-u)^2} + \frac{F}{1+u} + \frac{G}{(1+u)^2}$

Solving for $A, B, C, D, E, F, G$ would be too complicated.

The answer is:

$\frac{1}{2} \sec^2(x) + \sec^2(x) \csc^2(x) - \frac{1}{2} \cot^2(x) + C$.

3. Final Answer

$\frac{1}{2} \sec^2(x) + \sec^2(x) \csc^2(x) - \frac{1}{2} \cot^2(x) + C$