Integrals-U substitution for trig integrand

[jaredld]

The instructions are to evaluate the integral using any technique. Here is what I got.\(\displaystyle $\int {\sec ^2 t\,\tan t\;dt;\left[ \begin{array}{l}
u = \tan t \\
du = \sec ^2 dt \\
\end{array} \right]} \to - \ln |\cos u|\,du + C$\)

But this is the answer in the book. How do you get from what I have up top to the back of the book?
\(\displaystyle $\int {\ln |\sec u| + C \to \int {\ln |\sec (\tan t)| + C} } $\)

 

[soroban]

Hello, jaredld!

Didn't you see that you formatting didn't work?

How do you get from \(\displaystyle \,-\ln|\cos(\tan\,t)|\.\) to \(\displaystyle \.\ln|\sec(\tan\,t)|\) ?

Recall the properties of logarithms . . .

\(\displaystyle \,-\ln|\cos\theta|\;=\;\ln|\cos\theta|^{-1} \;=\;\ln\left|\frac{1}{|cos\theta}\right| \;= \;\ln|\sec\theta|\)

 

[jaredld]

thanks for the help. I should have seen it, but I didn't. Thanks for the reminder.

Yep I saw that it didn't work. I was in the process of correcting it. I didn't know enough about latex to know whether it would magically change the way it is posted once I submitted it.