1. ## Calc 2 Integration

I have to integrate

I'm getting
37((1/4(cos(2x)))-(log(cos(x))))

It is wrong though. Any help is appreciated.

Thanks,
Jay

2. the tan becomes sin/cos and that cancells off the adjacent $\displaystyle (cos{x})^2$. Then take out the 37 so now you've got $\displaystyle 37\int\frac{\sin{x}^3}{\cos{x}}dx$

3. still getting the same answer. It says to use ln(abs(u)) where appropriate
Would that make any difference?

4. Originally Posted by superdude
the tan becomes sin/cos and that cancells off the adjacent $\displaystyle (cos{x})^2$. Then take out the 37 so now you've got $\displaystyle 37\int\frac{\sin{x}^3}{\cos{x}}dx$
You forgot parentheses or didn't know how to write this is Latex. It should be $\displaystyle 37\int\frac{\sin^3{x}}{\cos{x}}dx$. Easy error to make.

Originally Posted by Hellacious D
still getting the same answer. It says to use ln(abs(u)) where appropriate
Would that make any difference?
Yes it would! Have you used u-substitution before? You should have if you are assigned this problem. Let u = cos(x) and rewrite the integral.