# Evaluate the integral. \int \sec^{2}x\tan x dx

Evaluate the integral.
$\int {\mathrm{sec}}^{2}x\mathrm{tan}xdx$
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Edward Patten

Step 1
Given: $I=\int {\mathrm{sec}}^{2}x\mathrm{tan}xdx$
for evaluating given integral, we substitute
$\mathrm{tan}x=t$...(1)
now, differentiating equation (1) with respect to x

${\mathrm{sec}}^{2}x=\frac{dt}{dx}$
${\mathrm{sec}}^{2}xdx=dt$
Step 2
now, replacing ${\mathrm{sec}}^{2}xdx$ with dt, tanx with t in given integral
so,

$=\left(\frac{{t}^{2}}{2}\right)+c$...(2)
now, replacing t with $\mathrm{tan}x$ in equation (2)
so,
$\int {\mathrm{sec}}^{2}x\mathrm{tan}xdx=\frac{{\mathrm{tan}}^{2}x}{2}+c$
hence, given integral is equal to $\frac{{\mathrm{tan}}^{2}x}{2}+c$.

###### Not exactly what you’re looking for?
Jack Maxson
$\int {\mathrm{sec}}^{2}\left(x\right)\mathrm{tan}\left(x\right)dx$
Substitution $u={\mathrm{sec}}^{2}\left(x\right)⇒\frac{du}{dx}=2{\mathrm{sec}}^{2}\left(x\right)\mathrm{tan}\left(x\right)⇒dx=\frac{1}{2{\mathrm{sec}}^{2}\left(x\right)\mathrm{tan}\left(x\right)}du:$
$=\frac{1}{2}\int 1du$
Now we calculate:
$\int 1du$
Integral of a constant:
=u
We substitute the already calculated integrals:
$\frac{1}{2}\int 1du$
$=\frac{u}{2}$
Reverse replacement $u={\mathrm{sec}}^{2}\left(x\right):$
$=\frac{{\mathrm{sec}}^{2}\left(x\right)}{2}$
$\int {\mathrm{sec}}^{2}\left(x\right)\mathrm{tan}\left(x\right)dx$
$=\frac{{\mathrm{sec}}^{2}\left(x\right)}{2}+C$
###### Not exactly what you’re looking for?
nick1337

$\int {\mathrm{sec}}^{2}\left(x\right)\mathrm{tan}\left(x\right)dx$
Apply u-substitution: $u=\mathrm{tan}\left(x\right)$
$=\int udu$
Apply the Power Rule
$=\frac{{u}^{2}}{2}$
Substitute back $u=\mathrm{tan}\left(x\right)$
$=\frac{{\mathrm{tan}}^{2}\left(x\right)}{2}$
Add a constant to the solution
$=\frac{{\mathrm{tan}}^{2}\left(x\right)}{2}+C$