Integral of $\sec^{3}{\left(x \right)}$

Find $\int \sec^{3}{\left(x \right)}\, dx$.

For the integral $\int{\sec^{3}{\left(x \right)} d x}$, use integration by parts $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$.

Let $\operatorname{u}=\sec{\left(x \right)}$ and $\operatorname{dv}=\sec^{2}{\left(x \right)} dx$.

Then $\operatorname{du}=\left(\sec{\left(x \right)}\right)^{\prime }dx=\tan{\left(x \right)} \sec{\left(x \right)} dx$ (steps can be seen ») and $\operatorname{v}=\int{\sec^{2}{\left(x \right)} d x}=\tan{\left(x \right)}$ (steps can be seen »).

The integral can be rewritten as

$$\int{\sec^{3}{\left(x \right)} d x}=\sec{\left(x \right)} \cdot \tan{\left(x \right)}-\int{\tan{\left(x \right)} \cdot \tan{\left(x \right)} \sec{\left(x \right)} d x}=\tan{\left(x \right)} \sec{\left(x \right)} - \int{\tan^{2}{\left(x \right)} \sec{\left(x \right)} d x}$$

Apply the formula $\tan^{2}{\left(x \right)} = \sec^{2}{\left(x \right)} - 1$:

$$\tan{\left(x \right)} \sec{\left(x \right)} - \int{\tan^{2}{\left(x \right)} \sec{\left(x \right)} d x}=\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec{\left(x \right)} d x}$$

Expand:

$$\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec{\left(x \right)} d x}=\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{3}{\left(x \right)} - \sec{\left(x \right)}\right)d x}$$

The integral of a difference is the difference of integrals:

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

Thus, we get the following simple linear equation with respect to the integral:

$${\color{red}{\int{\sec^{3}{\left(x \right)} d x}}}=\tan{\left(x \right)} \sec{\left(x \right)} + \int{\sec{\left(x \right)} d x} - {\color{red}{\int{\sec^{3}{\left(x \right)} d x}}}$$

Solving it, we obtain that

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

Rewrite the secant as $\sec\left(x\right)=\frac{1}{\cos\left(x\right)}$:

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

Rewrite the cosine in terms of the sine using the formula $\cos\left(x\right)=\sin\left(x + \frac{\pi}{2}\right)$ and then rewrite the sine using the double angle formula $\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$:

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

Multiply the numerator and denominator by $\sec^2\left(\frac{x}{2} + \frac{\pi}{4} \right)$:

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

Let $u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$.

Then $du=\left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2} dx$ (steps can be seen »), and we have that $\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} dx = 2 du$.

Therefore,

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

The integral of $\frac{1}{u}$ is $\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$:

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

Recall that $u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$:

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} + \frac{\tan{\left(x \right)} \sec{\left(x \right)}}{2} = \frac{\ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}}{2} + \frac{\tan{\left(x \right)} \sec{\left(x \right)}}{2}$$

Therefore,

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

Add the constant of integration:

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

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