We are asked to evaluate the integral $\int \frac{dx}{\sin^3(x) \cos^5(x)}$.

1. Problem Description

We are asked to evaluate the integral $\int \frac{dx}{\sin^3(x) \cos^5(x)}$.

2. Solution Steps

We can rewrite the integral using the identity $\sin^2(x) + \cos^2(x) = 1$:

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

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

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

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

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

We can write

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

Let $u = \tan(x)$, then $du = \sec^2(x) dx$.

We have $\sec^2(x) = 1 + \tan^2(x)$, so $\sec^8(x) = (\sec^2(x))^4 = (1+\tan^2(x))^4 = (1+u^2)^4$.

Then the integral becomes

$\int \frac{(1+u^2)^4}{u^3} du = \int \frac{1+4u^2+6u^4+4u^6+u^8}{u^3} du = \int (u^{-3} + 4u^{-1} + 6u + 4u^3 + u^5) du$.

$= \frac{u^{-2}}{-2} + 4\ln|u| + \frac{6u^2}{2} + \frac{4u^4}{4} + \frac{u^6}{6} + C = -\frac{1}{2u^2} + 4\ln|u| + 3u^2 + u^4 + \frac{u^6}{6} + C$.

$= -\frac{1}{2\tan^2(x)} + 4\ln|\tan(x)| + 3\tan^2(x) + \tan^4(x) + \frac{\tan^6(x)}{6} + C$.

Alternatively, we can rewrite the integral as:

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

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

The problem is still complicated.

Another way is to use the identity $1 = \cos^2(x) + \sin^2(x)$. Then $1 = (\cos^2(x) + \sin^2(x))^3 = \cos^6(x) + 3\cos^4(x)\sin^2(x) + 3\cos^2(x)\sin^4(x) + \sin^6(x)$.

Also $1 = (\cos^2(x) + \sin^2(x))^4 = \cos^8(x) + 4\cos^6(x)\sin^2(x) + 6\cos^4(x)\sin^4(x) + 4\cos^2(x)\sin^6(x) + \sin^8(x)$.

We can write

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

Let's go back to $\int \frac{\sec^8(x)}{\tan^3(x)} dx$.

$\int \frac{\sec^8(x)}{\tan^3(x)} dx = \int \frac{\sec^6(x) \sec^2(x)}{\tan^3(x)} dx = \int \frac{(1+\tan^2(x))^3 \sec^2(x)}{\tan^3(x)} dx$.

Using $u = \tan(x)$, we have $\int \frac{(1+u^2)^3}{u^3} du = \int \frac{1+3u^2+3u^4+u^6}{u^3} du = \int (u^{-3} + 3u^{-1} + 3u + u^3) du$

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

3. Final Answer

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