Integrating secant squared times tangent Integrate the function.
$$ \int \sec^2 x \tan x dx $$
I'm trying to get a proper substitution, but I couldn't get anything proper.
 A: We know that,
$$ d(\tan(x)) = \sec^2(x).dx$$
Plugging the LHS of the above equation in place of RHS into the main expression gives:
$$ \int \tan x. d( \tan x )$$
$$ \frac{\tan^2(x)}{2} + Constant$$
A: This is a good one. Hope you are familiar with integration by substitution.
now case 1:
put $\\ \tan x=t$ so, $ dt=\sec^2xdx $
Substituting, we get 
$I= \int tdt\Rightarrow \frac {t^2}{2}+c_1 $
that is, I= $ \frac{\tan^2x}{2} + c_1 $
Case 2: put $ \sec x=u $
$du=\sec x\tan x \ dx$
so, $I= \int udu => I=\frac {u^2}{2}+c_2$
since both the integrals are equal,
$\frac{\tan^2x}{2}+c_1=\frac {\sec^2x}{2}+c_2 $
which implies $c_1 -c_2 =\frac {1}{2} $.
So, we understand that, if a function has more than one integral, the integrals differ by a constant.
A: Substitute $u=\tan x$, then you'll get $du=\sec^2(x) dx$
$$\int \sec^2(x) \cdot \tan x dx = \int u \; du = \frac{u^2}{2}+c \Rightarrow \frac{\tan^2(x)}{2}+c$$
Now we use $\sec^2(x)=1+\tan^2(x)$ and finally
$$\int \sec^2x \cdot \tan x dx = \frac{\sec^2(x)}{2}+c$$
