Integral of $$$\sec^{4}{\left(\theta \right)}$$$
Related calculator: Definite and Improper Integral Calculator
Your Input
Find $$$\int \sec^{4}{\left(\theta \right)}\, d\theta$$$.
Solution
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 $$$\alpha=\theta$$$:
$${\color{red}{\int{\sec^{4}{\left(\theta \right)} d \theta}}} = {\color{red}{\int{\left(\tan^{2}{\left(\theta \right)} + 1\right) \sec^{2}{\left(\theta \right)} d \theta}}}$$
Let $$$u=\tan{\left(\theta \right)}$$$.
Then $$$du=\left(\tan{\left(\theta \right)}\right)^{\prime }d\theta = \sec^{2}{\left(\theta \right)} d\theta$$$ (steps can be seen »), and we have that $$$\sec^{2}{\left(\theta \right)} d\theta = du$$$.
Therefore,
$${\color{red}{\int{\left(\tan^{2}{\left(\theta \right)} + 1\right) \sec^{2}{\left(\theta \right)} d \theta}}} = {\color{red}{\int{\left(u^{2} + 1\right)d u}}}$$
Integrate term by term:
$${\color{red}{\int{\left(u^{2} + 1\right)d u}}} = {\color{red}{\left(\int{1 d u} + \int{u^{2} d u}\right)}}$$
Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:
$$\int{u^{2} d u} + {\color{red}{\int{1 d u}}} = \int{u^{2} d u} + {\color{red}{u}}$$
Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:
$$u + {\color{red}{\int{u^{2} d u}}}=u + {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=u + {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$
Recall that $$$u=\tan{\left(\theta \right)}$$$:
$${\color{red}{u}} + \frac{{\color{red}{u}}^{3}}{3} = {\color{red}{\tan{\left(\theta \right)}}} + \frac{{\color{red}{\tan{\left(\theta \right)}}}^{3}}{3}$$
Therefore,
$$\int{\sec^{4}{\left(\theta \right)} d \theta} = \frac{\tan^{3}{\left(\theta \right)}}{3} + \tan{\left(\theta \right)}$$
Add the constant of integration:
$$\int{\sec^{4}{\left(\theta \right)} d \theta} = \frac{\tan^{3}{\left(\theta \right)}}{3} + \tan{\left(\theta \right)}+C$$
Answer
$$$\int \sec^{4}{\left(\theta \right)}\, d\theta = \left(\frac{\tan^{3}{\left(\theta \right)}}{3} + \tan{\left(\theta \right)}\right) + C$$$A