# Find the integral: int sec^3 xdx

Find the integral:
$\int {\mathrm{sec}}^{3}xdx$
You can still ask an expert for help

## Want to know more about Derivatives?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Leo Robinson
Use Integration by Parts on $\int {\mathrm{sec}}^{3}xdx$
Let $u=\mathrm{sec}x,dv={\mathrm{sec}}^{2}x,du=\mathrm{sec}x\mathrm{tan}xdx,v=\mathrm{tan}x$
Substitute the above into $uv-\int vdu.$
$\mathrm{sec}x\mathrm{tan}x-\int {\mathrm{tan}}^{2}x\mathrm{sec}xdx$
Use Pythagorean Identities: ${\mathrm{tan}}^{2}x={\mathrm{sec}}^{2}x-1$
$\mathrm{sec}x\mathrm{tan}x-\int \left({\mathrm{sec}}^{2}x-1\right)\mathrm{sec}xdx$
Expand.
$\mathrm{sec}x\mathrm{tan}x-\int {\mathrm{sec}}^{3}x-\mathrm{sec}xdx$
Use Sum Rule:$\int f\left(x\right)+g\left(x\right)dx=\int f\left(x\right)dx+\int g\left(x\right)dx.$
$\mathrm{sec}x\mathrm{tan}x-\int {\mathrm{sec}}^{3}dx+\int \mathrm{sec}xdx$
Set it as equal to the original integral $\int {\mathrm{sec}}^{3}xdx$
$\int {\mathrm{sec}}^{3}xdx=\mathrm{sec}x\mathrm{tan}x-\int {\mathrm{sec}}^{3}xdx+\int \mathrm{sec}xdx$
Add $\int {\mathrm{sec}}^{3}xdx$ to both sides.
$\int {\mathrm{sec}}^{3}xdx+\int {\mathrm{sec}}^{3}xdx=\mathrm{sec}x\mathrm{tan}x+\int \mathrm{sec}xdx$
Simplify $\int {\mathrm{sec}}^{3}xdx+\int {\mathrm{sec}}^{3}xdx$ to $2\int {\mathrm{sec}}^{3}xdx$
$2\int {\mathrm{sec}}^{3}xdx=\mathrm{sec}x\mathrm{tan}x+\int \mathrm{sec}xdx$
Divide both sides by $2$
$\int {\mathrm{sec}}^{3}xdx=\frac{\mathrm{sec}x\mathrm{tan}x+\int \mathrm{sec}xdx}{2}$
Original integral solved.
$\frac{\mathrm{sec}x\mathrm{tan}x+\int \mathrm{sec}xdx}{2}$
Use Trigonometric Integration: the integral of $\mathrm{sec}x$
$\mathrm{ln}\left(\mathrm{sec}x+\mathrm{tan}x\right)$
$\frac{\mathrm{sec}x\mathrm{tan}x+\mathrm{ln}\left(\mathrm{sec}x+\mathrm{tan}x\right)}{2}$
$\frac{\mathrm{sec}x\mathrm{tan}x+\mathrm{ln}\left(\mathrm{sec}x+\mathrm{tan}x\right)}{2}+C$