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

$$\int x\sec^2 xdx=x\tan x+\ln|\cos x|+C$$
$$A=\int x\sec^2 xdx$$ Take $u=x$ and $dv=\sec^2xdx$ We then have $du=dx$ and $v=\tan x$ Apply the formula $\int udv= uv-\int vdu$, we have $$A=x\tan x-\int\tan xdx$$ $$A=x\tan x-\int\frac{\sin x}{\cos x}dx$$ Take $z=\cos x$, we then have $dz=-\sin xdx$, meaning that $\sin xdx=-dz$ That means $$A=x\tan x-\int\frac{-dz}{z}$$ $$A=x\tan x+\int\frac{1}{z}dz$$ $$A=x\tan x+\ln|z|+C$$ Now replace back $z$ with $\cos x$: $$A=x\tan x+\ln|\cos x|+C$$