Integral of $$$\frac{1}{\sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)}}$$$

Find $$$\int \frac{1}{\sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)}}\, dx$$$.

Multiply the numerator and denominator by $$$\frac{1}{\cos^{4}{\left(x \right)}}$$$ and convert $$$\frac{\cos^{4}{\left(x \right)}}{\sin^{4}{\left(x \right)}}$$$ into $$$\frac{1}{\tan^{4}{\left(x \right)}}$$$:

$${\color{red}{\int{\frac{1}{\sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{\cos^{8}{\left(x \right)} \tan^{4}{\left(x \right)}} d x}}}$$

Strip out two cosines and rewrite them in terms of the secant using the formula $$$\frac{1}{\cos^{2}{\left(x \right)}}=\sec^{2}{\left(x \right)}$$$:

$${\color{red}{\int{\frac{1}{\cos^{8}{\left(x \right)} \tan^{4}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\sec^{2}{\left(x \right)}}{\cos^{6}{\left(x \right)} \tan^{4}{\left(x \right)}} d x}}}$$

Rewrite the cosine in terms of the tangent using the formula $$$\cos^{2}{\left(x \right)}=\frac{1}{\tan^{2}{\left(x \right)} + 1}$$$:

$${\color{red}{\int{\frac{\sec^{2}{\left(x \right)}}{\cos^{6}{\left(x \right)} \tan^{4}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\left(\tan^{2}{\left(x \right)} + 1\right)^{3} \sec^{2}{\left(x \right)}}{\tan^{4}{\left(x \right)}} d x}}}$$

Let $$$u=\tan{\left(x \right)}$$$.

Then $$$du=\left(\tan{\left(x \right)}\right)^{\prime }dx = \sec^{2}{\left(x \right)} dx$$$ (steps can be seen »), and we have that $$$\sec^{2}{\left(x \right)} dx = du$$$.

The integral can be rewritten as

$${\color{red}{\int{\frac{\left(\tan^{2}{\left(x \right)} + 1\right)^{3} \sec^{2}{\left(x \right)}}{\tan^{4}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\left(u^{2} + 1\right)^{3}}{u^{4}} d u}}}$$

Expand the expression:

$${\color{red}{\int{\frac{\left(u^{2} + 1\right)^{3}}{u^{4}} d u}}} = {\color{red}{\int{\left(u^{2} + 3 + \frac{3}{u^{2}} + \frac{1}{u^{4}}\right)d u}}}$$

Integrate term by term:

$${\color{red}{\int{\left(u^{2} + 3 + \frac{3}{u^{2}} + \frac{1}{u^{4}}\right)d u}}} = {\color{red}{\left(\int{3 d u} + \int{\frac{1}{u^{4}} d u} + \int{\frac{3}{u^{2}} d u} + \int{u^{2} d u}\right)}}$$

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=3$$$:

$$\int{\frac{1}{u^{4}} d u} + \int{\frac{3}{u^{2}} d u} + \int{u^{2} d u} + {\color{red}{\int{3 d u}}} = \int{\frac{1}{u^{4}} d u} + \int{\frac{3}{u^{2}} d u} + \int{u^{2} d u} + {\color{red}{\left(3 u\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-4$$$:

$$3 u + \int{\frac{3}{u^{2}} d u} + \int{u^{2} d u} + {\color{red}{\int{\frac{1}{u^{4}} d u}}}=3 u + \int{\frac{3}{u^{2}} d u} + \int{u^{2} d u} + {\color{red}{\int{u^{-4} d u}}}=3 u + \int{\frac{3}{u^{2}} d u} + \int{u^{2} d u} + {\color{red}{\frac{u^{-4 + 1}}{-4 + 1}}}=3 u + \int{\frac{3}{u^{2}} d u} + \int{u^{2} d u} + {\color{red}{\left(- \frac{u^{-3}}{3}\right)}}=3 u + \int{\frac{3}{u^{2}} d u} + \int{u^{2} d u} + {\color{red}{\left(- \frac{1}{3 u^{3}}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

$$3 u + \int{\frac{3}{u^{2}} d u} + {\color{red}{\int{u^{2} d u}}} - \frac{1}{3 u^{3}}=3 u + \int{\frac{3}{u^{2}} d u} + {\color{red}{\frac{u^{1 + 2}}{1 + 2}}} - \frac{1}{3 u^{3}}=3 u + \int{\frac{3}{u^{2}} d u} + {\color{red}{\left(\frac{u^{3}}{3}\right)}} - \frac{1}{3 u^{3}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=3$$$ and $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$:

$$\frac{u^{3}}{3} + 3 u + {\color{red}{\int{\frac{3}{u^{2}} d u}}} - \frac{1}{3 u^{3}} = \frac{u^{3}}{3} + 3 u + {\color{red}{\left(3 \int{\frac{1}{u^{2}} d u}\right)}} - \frac{1}{3 u^{3}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-2$$$:

$$\frac{u^{3}}{3} + 3 u + 3 {\color{red}{\int{\frac{1}{u^{2}} d u}}} - \frac{1}{3 u^{3}}=\frac{u^{3}}{3} + 3 u + 3 {\color{red}{\int{u^{-2} d u}}} - \frac{1}{3 u^{3}}=\frac{u^{3}}{3} + 3 u + 3 {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}} - \frac{1}{3 u^{3}}=\frac{u^{3}}{3} + 3 u + 3 {\color{red}{\left(- u^{-1}\right)}} - \frac{1}{3 u^{3}}=\frac{u^{3}}{3} + 3 u + 3 {\color{red}{\left(- \frac{1}{u}\right)}} - \frac{1}{3 u^{3}}$$

Recall that $$$u=\tan{\left(x \right)}$$$:

$$- \frac{{\color{red}{u}}^{-3}}{3} - 3 {\color{red}{u}}^{-1} + 3 {\color{red}{u}} + \frac{{\color{red}{u}}^{3}}{3} = - \frac{{\color{red}{\tan{\left(x \right)}}}^{-3}}{3} - 3 {\color{red}{\tan{\left(x \right)}}}^{-1} + 3 {\color{red}{\tan{\left(x \right)}}} + \frac{{\color{red}{\tan{\left(x \right)}}}^{3}}{3}$$

Therefore,

$$\int{\frac{1}{\sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)}} d x} = \frac{\tan^{3}{\left(x \right)}}{3} + 3 \tan{\left(x \right)} - \frac{3}{\tan{\left(x \right)}} - \frac{1}{3 \tan^{3}{\left(x \right)}}$$

Add the constant of integration:

$$\int{\frac{1}{\sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)}} d x} = \frac{\tan^{3}{\left(x \right)}}{3} + 3 \tan{\left(x \right)} - \frac{3}{\tan{\left(x \right)}} - \frac{1}{3 \tan^{3}{\left(x \right)}}+C$$

$$$\int \frac{1}{\sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)}}\, dx = \left(\frac{\tan^{3}{\left(x \right)}}{3} + 3 \tan{\left(x \right)} - \frac{3}{\tan{\left(x \right)}} - \frac{1}{3 \tan^{3}{\left(x \right)}}\right) + C$$$A