Integral of $$$\sec^{5}{\left(u \right)}$$$

Find $$$\int \sec^{5}{\left(u \right)}\, du$$$.

For the integral $$$\int{\sec^{5}{\left(u \right)} d u}$$$, use integration by parts $$$\int \operatorname{o} \operatorname{dv} = \operatorname{o}\operatorname{v} - \int \operatorname{v} \operatorname{do}$$$.

Let $$$\operatorname{o}=\sec^{3}{\left(u \right)}$$$ and $$$\operatorname{dv}=\sec^{2}{\left(u \right)} du$$$.

Then $$$\operatorname{do}=\left(\sec^{3}{\left(u \right)}\right)^{\prime }du=3 \tan{\left(u \right)} \sec^{3}{\left(u \right)} du$$$ (steps can be seen ») and $$$\operatorname{v}=\int{\sec^{2}{\left(u \right)} d u}=\tan{\left(u \right)}$$$ (steps can be seen »).

So,

$$\int{\sec^{5}{\left(u \right)} d u}=\sec^{3}{\left(u \right)} \cdot \tan{\left(u \right)}-\int{\tan{\left(u \right)} \cdot 3 \tan{\left(u \right)} \sec^{3}{\left(u \right)} d u}=\tan{\left(u \right)} \sec^{3}{\left(u \right)} - \int{3 \tan^{2}{\left(u \right)} \sec^{3}{\left(u \right)} d u}$$

Strip out the constant:

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

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

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

Expand:

$$\tan{\left(u \right)} \sec^{3}{\left(u \right)} - 3 \int{\left(\sec^{2}{\left(u \right)} - 1\right) \sec^{3}{\left(u \right)} d u}=\tan{\left(u \right)} \sec^{3}{\left(u \right)} - 3 \int{\left(\sec^{5}{\left(u \right)} - \sec^{3}{\left(u \right)}\right)d u}$$

The integral of a difference is the difference of integrals:

$$\tan{\left(u \right)} \sec^{3}{\left(u \right)} - 3 \int{\left(\sec^{5}{\left(u \right)} - \sec^{3}{\left(u \right)}\right)d u}=\tan{\left(u \right)} \sec^{3}{\left(u \right)} + 3 \int{\sec^{3}{\left(u \right)} d u} - 3 \int{\sec^{5}{\left(u \right)} d u}$$

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

$${\color{red}{\int{\sec^{5}{\left(u \right)} d u}}}=\tan{\left(u \right)} \sec^{3}{\left(u \right)} + 3 \int{\sec^{3}{\left(u \right)} d u} - 3 {\color{red}{\int{\sec^{5}{\left(u \right)} d u}}}$$

Solving it, we obtain that

$$\int{\sec^{5}{\left(u \right)} d u}=\frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \int{\sec^{3}{\left(u \right)} d u}}{4}$$

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

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

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

So,

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

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

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

Expand:

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

The integral of a difference is the difference of integrals:

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

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

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

Solving it, we obtain that

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

Therefore,

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

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

$$\frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\sec{\left(u \right)} d u}}}}{8} = \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{\cos{\left(u \right)}} d u}}}}{8}$$

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

$$\frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{\cos{\left(u \right)}} d u}}}}{8} = \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{u}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{8}$$

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

$$\frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{u}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{8} = \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{8}$$

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

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

So,

$$\frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{8} = \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{v} d v}}}}{8}$$

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

$$\frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{v} d v}}}}{8} = \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{8}$$

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

$$\frac{3 \ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{8} + \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} = \frac{3 \ln{\left(\left|{{\color{red}{\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}}{8} + \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8}$$

Therefore,

$$\int{\sec^{5}{\left(u \right)} d u} = \frac{3 \ln{\left(\left|{\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{8} + \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8}$$

Add the constant of integration:

$$\int{\sec^{5}{\left(u \right)} d u} = \frac{3 \ln{\left(\left|{\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{8} + \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8}+C$$

$$$\int \sec^{5}{\left(u \right)}\, du = \left(\frac{3 \ln\left(\left|{\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}\right|\right)}{8} + \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8}\right) + C$$$A