## University Calculus: Early Transcendentals (3rd Edition)

$\frac{1}{2}tan^{2}x+C$
Substitute $u= tan x$ so that $dx=\frac{du}{sec^{2}x}$ in the given integral. Then, $\int sec^{2}xtanxdx=\int sec^{2}x\, u\frac{du}{sec^{2}x}$ $=\int udu= \frac{u^{2}}{2}+C$ Undoing substitution, we get $\int sec^{2}x tanxdx= \frac{1}{2}tan^{2}x+C$