Hi
Evaluate the integral of t(sec^2(6t))dt
My answer is: (t/6)tan(6t) - (1/12)Ln|sec(6t)| + C
why is this wrong?
Thanks
Hello, sjmiller!
Integrate: .$\displaystyle I \;=\;\int t\sec^2(6t)\,dt$
My answer is: .$\displaystyle \tfrac{1}{6}t\tan(6t) - \tfrac{1}{12}\ln|\sec(6t)| + C$
Why is this wrong?
Did you do something silly? . . . .Like: $\displaystyle \tfrac{1}{6}\!\cdot\!\tfrac{1}{6} \,=\,\tfrac{1}{12}$ ??
By parts: .$\displaystyle \begin{Bmatrix}u &=& t && dv &=& \sec^2(6t)\,dt \\ du &=& dt && v &=& \frac{1}{6}\tan(6t) \end{Bmatrix}$
We have: .$\displaystyle I \;=\;\tfrac{1}{6}t\tan(6t) - \tfrac{1}{6}\int\tan(6t)\,dt$
. . . . . . . . $\displaystyle I \;=\;\tfrac{1}{6}t\tan(6t) - {\bf\tfrac{1}{36}}\ln|\sec(6t)| + C$
. . . . . . . . . . . . . . . . . . . . . $\displaystyle \uparrow$