Hey Mr. Sjoberg: What did I do wrong here? Section 3.2, Ex. 45: Evaluate: $\int \sec^2{x}\tan{x} \, dx$ $$\implies \int \sec{x}(\sec{x}\tan{x})\, dx$$ $$\text{let } u = \sec x$$ $$\implies \mathrm{d}u = \sec{x}\tan{x}\, dx$$ $$\implies \int u \, du$$ $$ = \frac{1}{2} u^2 + C$$ $$ = \frac{1}{2} \sec^2{x} + C$$ \medskip But the back of the book says $\dfrac{1}{2}\tan^2{x} + C$. What's up with that?