Integral of $$$64 \sec^{4}{\left(x \right)}$$$

Find $$$\int 64 \sec^{4}{\left(x \right)}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=64$$$ and $$$f{\left(x \right)} = \sec^{4}{\left(x \right)}$$$:

$${\color{red}{\int{64 \sec^{4}{\left(x \right)} d x}}} = {\color{red}{\left(64 \int{\sec^{4}{\left(x \right)} d x}\right)}}$$

Strip out two secants and write everything else in terms of the tangent, using the formula $$$\sec^2\left( \alpha \right)=\tan^2\left( \alpha \right) + 1$$$ with $$$\alpha=x$$$:

$$64 {\color{red}{\int{\sec^{4}{\left(x \right)} d x}}} = 64 {\color{red}{\int{\left(\tan^{2}{\left(x \right)} + 1\right) \sec^{2}{\left(x \right)} d x}}}$$

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

$$64 {\color{red}{\int{\left(\tan^{2}{\left(x \right)} + 1\right) \sec^{2}{\left(x \right)} d x}}} = 64 {\color{red}{\int{\left(u^{2} + 1\right)d u}}}$$

Integrate term by term:

$$64 {\color{red}{\int{\left(u^{2} + 1\right)d u}}} = 64 {\color{red}{\left(\int{1 d u} + \int{u^{2} d u}\right)}}$$

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

$$64 \int{u^{2} d u} + 64 {\color{red}{\int{1 d u}}} = 64 \int{u^{2} d u} + 64 {\color{red}{u}}$$

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

$$64 u + 64 {\color{red}{\int{u^{2} d u}}}=64 u + 64 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=64 u + 64 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

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

$$64 {\color{red}{u}} + \frac{64 {\color{red}{u}}^{3}}{3} = 64 {\color{red}{\tan{\left(x \right)}}} + \frac{64 {\color{red}{\tan{\left(x \right)}}}^{3}}{3}$$

Therefore,

$$\int{64 \sec^{4}{\left(x \right)} d x} = \frac{64 \tan^{3}{\left(x \right)}}{3} + 64 \tan{\left(x \right)}$$

Simplify:

$$\int{64 \sec^{4}{\left(x \right)} d x} = \frac{64 \left(\tan^{2}{\left(x \right)} + 3\right) \tan{\left(x \right)}}{3}$$

Add the constant of integration:

$$\int{64 \sec^{4}{\left(x \right)} d x} = \frac{64 \left(\tan^{2}{\left(x \right)} + 3\right) \tan{\left(x \right)}}{3}+C$$

$$$\int 64 \sec^{4}{\left(x \right)}\, dx = \frac{64 \left(\tan^{2}{\left(x \right)} + 3\right) \tan{\left(x \right)}}{3} + C$$$A