more integration

[legacyofpiracy]

first off would you suggest I use integration by parts in this problem

 S x sin(2x)dx

I was using u=x and dv=sin(2x) but nothing was going to well so perhaps my choices were bad?

the same thing for

(e^(tanx))/(cos^2)x

In this one I tried using tanx for u, but I was having trouble getting the cos^2 to fit in there.

 

[soroban]

Hello, legacyofpiracy!

\(\displaystyle \L \int x\cdot\sin(2x)\,dx\)

I was using \(\displaystyle u\,=\,x\) and \(\displaystyle dv\,=\,sin(2x)\,dx\), but nothing was going too well

This should have worked . . .

Let: \(\displaystyle u\,=\,x\;\;\;\;\;\;dv\,=\,\sin(2x)\,dx\)

Then: \(\displaystyle du\,=\,dx\;\;\;v\,=\,-\frac{1}{2}\cos(2x)\)

And we have: \(\displaystyle \L\,-\frac{1}{2}x\cdot\cos(2x) \,+ \,\frac{1}{2}\int \cos(2x)\,dx\)

I don't see any problem in finishing it . . .

\(\displaystyle \L \int \frac{e^{^{\tan x}}}{\cos^2x}\,dx\)

We have: \(\displaystyle \L\:\int e^{^{\tan x}}(\sec^2x)\,dx\)

Let: \(\displaystyle u \:=\:\tan x\)

 

[legacyofpiracy]

thank you soroban, just a question. On the first one would I have to continue on and find another value of u, v, etc. or can I just integrate it and leave as is? In otherwords could i simply leave it as

-1/2 xcos(2x)+ 1/2 sin(2x)*2+C ?

 

[Gene]

Close but
d(sin(2x))=2cos(2x)dx
Think about it.

 

[legacyofpiracy]

I'm sorry, i'm not following. Which sin(2x) are you referring to? If its the final one how can i take the derivative of it?

 

[Gene]

tsk tsk tsk!
d(sin(2x))=2cos(2x)dx
Int(d(sin(2x)))=int(2cos(2x))dx
sin(2x) = int(2cos(2x))dx
so
(1/2)int(cos(2x))dx = (1/4)int(2cos(2x))dx =
(1/4)sin(2x)