Integration by Parts

Hello, sjmiller!

Quote:

Integrate: . $I \;=\;\int t\sec^2(6t)\,dt$

My answer is: . $\tfrac{1}{6}t\tan(6t) - \tfrac{1}{12}\ln|\sec(6t)| + C$

Why is this wrong?

Did you do something silly? . . . .Like: $\tfrac{1}{6}\!\cdot\!\tfrac{1}{6} \,=\,\tfrac{1}{12}$ ??

By parts: . $\begin{Bmatrix}u &=& t && dv &=& \sec^2(6t)\,dt \\ du &=& dt && v &=& \frac{1}{6}\tan(6t) \end{Bmatrix}$

We have: . $I \;=\;\tfrac{1}{6}t\tan(6t) - \tfrac{1}{6}\int\tan(6t)\,dt$

. . . . . . . . $I \;=\;\tfrac{1}{6}t\tan(6t) - {\bf\tfrac{1}{36}}\ln|\sec(6t)| + C$
. . . . . . . . . . . . . . . . . . . . . $\uparrow$