We need to find the integral of the following expression: $\int \frac{dx}{\sin^3(x) \cos^5(x)}$

1. Problem Description

We need to find the integral of the following expression:

$\int \frac{dx}{\sin^3(x) \cos^5(x)}$

2. Solution Steps

We can rewrite the integral as follows:

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

We can rewrite $1$ as $(\sin^2(x) + \cos^2(x))^n$ for any $n$.

Let's rewrite the integral as:

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

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

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

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

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

Let us rewrite the original integral as:

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

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

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

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

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

We have $\int \frac{dx}{\sin^3(x) \cos^5(x)} = \int \frac{dx}{\sin^3(x) \cos^5(x)} \cdot \frac{\cos^2(x) + \sin^2(x)}{1} = \int \frac{\sin^2(x) + \cos^2(x)}{\sin^3(x) \cos^5(x)}dx = \int \frac{\sin^2(x)}{\sin^3(x) \cos^5(x)} + \frac{\cos^2(x)}{\sin^3(x) \cos^5(x)}dx = \int \frac{dx}{\sin(x)\cos^5(x)} + \int \frac{dx}{\sin^3(x)\cos^3(x)}$

Now, consider $\int \frac{dx}{\sin^3(x)\cos^5(x)} = \int \frac{\sin^2(x) + \cos^2(x)}{\sin^3(x)\cos^5(x)}dx = \int \frac{1}{\sin(x)\cos^5(x)}dx + \int \frac{1}{\sin^3(x)\cos^3(x)}dx$

$\int \frac{dx}{\sin^3(x) \cos^5(x)} = \int \frac{1}{\sin^3(x) \cos^5(x)} dx = \int \frac{1}{\frac{\sin^3(x)}{\cos^3(x)} \cos^8(x)} dx = \int \frac{\sec^8(x)}{\tan^3(x)} dx = \int \frac{\sec^6(x)}{\tan^3(x)} \sec^2(x) dx$

Now let $u = \tan(x)$, so $du = \sec^2(x) dx$. Also $\sec^2(x) = 1 + \tan^2(x) = 1 + u^2$

So $\sec^6(x) = (1 + u^2)^3 = 1 + 3u^2 + 3u^4 + u^6$

Thus we have $\int \frac{1 + 3u^2 + 3u^4 + u^6}{u^3} du = \int (\frac{1}{u^3} + \frac{3}{u} + 3u + u^3) du$

$= \frac{-1}{2u^2} + 3\ln|u| + \frac{3u^2}{2} + \frac{u^4}{4} + C$

$= \frac{-1}{2\tan^2(x)} + 3\ln|\tan(x)| + \frac{3\tan^2(x)}{2} + \frac{\tan^4(x)}{4} + C$

3. Final Answer

$\frac{\tan^4(x)}{4} + \frac{3\tan^2(x)}{2} + 3\ln|\tan(x)| - \frac{1}{2\tan^2(x)} + C$