Help starting two integrals

Oct 2008
13
0
The first one is: The integral of e^(2sinx) / (5secx) so far I have tried do a substitution and letting u = 2sinx so du = 2cosxdx. I'm pretty sure you can move 5secx into the numerator, making it 5cosx. But from there, I'm not sure if I should move the 2 to the other side, making it (1/2)du, or if I can cancel out 2cosx. Also I'm not really sure what else to do...since it would then be the integral of e^(u) * 5cosx du? Is it possible to do it with an x and a u?

The second one is the integral of ((4 + e^ (-3x)) * (e^(2x)). For this one, would you multiply it out so you can get the integral of (4e^(2x)) + the integral of ((e^(-3x)*(e^(2x))? I'm not sure how to deal with e when it is raised to two different things.

Any help would be great and thank you!
 

pickslides

MHF Helper
Sep 2008
5,237
1,625
Melbourne
\(\displaystyle \int \frac{e^{\sin x}}{5 \sec x} ~dx\)

\(\displaystyle \frac{1}{5}\int \frac{e^{\sin x}}{ \sec x} ~dx\)

\(\displaystyle \frac{1}{5}\int e^{\sin x} \cos x ~dx\)

\(\displaystyle \frac{1}{5}\int e^{u} \frac{du}{dx} ~dx\)

\(\displaystyle \frac{1}{5}\int e^{u} ~du\)

\(\displaystyle \frac{1}{5} e^{u}+C\)

\(\displaystyle \frac{1}{5} e^{\sin x}+C\)
 
Oct 2008
13
0
Oh wow, I'm not sure how I didn't see that, but thank you very much! It's been a year and a half since I took calculus one, so I'm a little rusty. Thank you, appreciate it!
 

pickslides

MHF Helper
Sep 2008
5,237
1,625
Melbourne
((e^(-3x)*(e^(2x))? I'm not sure how to deal with e when it is raised to two different things.
\(\displaystyle a^m\times a^n = a^{m+n} \)

therefore

\(\displaystyle e^{-3x}\times e^{2x} = e^{-3x+2x} \)