hi, please help me with a question:
integrate $\displaystyle sec^3x tanx$
i simplified the expression to this: $\displaystyle tanx sec^2x secx$
i don't know how to integrate the above expression though
thank you in advance
Rewrite as
$\displaystyle \int{\sec^2{x}\sec{x}\tan{x}\,dx}$.
Let $\displaystyle u = \sec{x}$ so that $\displaystyle \frac{du}{dx} = \sec{x}\tan{x}$.
So the integral becomes
$\displaystyle \int{u^2\,\frac{du}{dx}\,dx}$
$\displaystyle = \int{u^2\,du}$
$\displaystyle = \frac{1}{3}u^3 + C$
$\displaystyle = \frac{1}{3}\sec^3{x} + C$.