Finding $\int \sec^3x\,dx$

Solution 1:

$$ \begin{align} &\int \sec^3x\,dx\\ =&\int \sec x \cdot \sec^2x\,dx\\ \end{align} $$ Parts: $u = \sec x,\, dv = \sec^2x\,dx$ $$ \begin{align} &\sec x \tan x - \int \sec x \tan^2x\,dx\\ =\,&\sec x \tan x - \int \sec x\left(\sec^2x-1\right)dx\\ =\,&\sec x \tan x - \int \sec^3x\,dx + \int\sec x\,dx \\ \end{align} $$ Using $\int\sec x\,dx = \ln\left|\sec x + \tan x\right| + C$, we have: $$ \begin{align} \int \sec^3x\,dx &=\sec x \tan x - \int \sec^3x\,dx + \ln\left|\sec x + \tan x\right| + C\\ 2\int \sec^3x\,dx &=\sec x \tan x + \ln\left|\sec x + \tan x\right| + C\\ \int \sec^3x\,dx &=\frac{1}{2}\sec x \tan x + \frac{1}{2}\ln\left|\sec x + \tan x\right| + C\\ \end{align} $$

This is just one way of solving it! This is a notoriously tricky integral. You can also do this with partial fractions, if you know those. I did this solution first because it sticks to more elementary calculus knowledge.

Solution 2:

Notice, $$\int \sec^3x dx$$$$=\int \sec x \sec^2 x dx$$ $$=\int \sqrt{1+\tan^2 x} \sec^2 x dx$$ Let, $\tan x=t\implies \sec^2x dx=dt$

$$=\int \sqrt{1+t^2}dt$$ $$=\frac{1}{2}\left(t\sqrt{1+t^2}+\ln\left|t+\sqrt{1+t^2}\right|\right)+C$$ $$=\frac{1}{2}\left(\tan x\sec x+\ln\left|\tan x +\sec x\right|\right)+C$$

Solution 3:

A less simple (but not complicated) solution using the tangent half-angle substitution $t=\tan(\frac x2)$. $$I=\int \sec^3x\,dx=\int\frac{2 \left(t^2+1\right)^2}{\left(1-t^2\right)^3}\,dt$$ Now, partial fraction decomposition $$\frac{2 \left(t^2+1\right)^2}{\left(1-t^2\right)^3}=\frac{1}{2 (t+1)}-\frac{1}{2 (t+1)^2}+\frac{1}{(t+1)^3}-\frac{1}{2 (t-1)}-\frac{1}{2 (t-1)^2}-\frac{1}{(t-1)^3}$$ Integrating and simplifying $$I=\frac{t^3+t}{\left(t^2-1\right)^2}+\frac 12 \log\Big(\frac{1+t}{1-t}\Big)$$