U Substitution: Solve \int sec^3(2x)tan(2x) - Casey

[Saladsamurai]

So I have another U substitution.

[tex]\int sec^3(2x)tan(2x)[/tex] this one is a little tricky for me. I have tried letting u= sec2x and tanx and 2x.
2x definitley gets me nowhere. I may be mistaken on the others. I will recheck them.

I was also thinking of rewriting it as
[tex]\int sec^4(2x)sin(2x)[/tex]

is the latter the better option?

Thanks,
Casey

 

[mattmns]

Unless I am mistaken, u = sec2x should work.

 

[Saladsamurai]

Oh...one second, let me re-work this.

 

[Saladsamurai]

Unless I am mistaken, u = sec2x should work.

I must be missing something...

[tex]\int sec^32xtan2x[/tex] If u=sec2x du=2sec2xtan2x dx
[tex]\rightarrow \int u^3 tan2x[/tex]

wait...how about

[tex]\int sec2xtanx*u^2 dx[/tex]
=[tex]\frac{1}{2}\int u^2du[/tex]

Thanks...Btw, how do you make a new line with Latex? I though it was \\

 

[rocomath]

your last 2 lines look funny to me, from what i have written down it should look like

[tex]\int\sec^{2}{2x}\sec{2x}\tan{2x}dx[/tex]

subst. [tex]\sec{2x}tan{2x}dx[/tex] with [tex]du[/tex]

[tex]\frac{1}{2} \int u^{2}du[/tex]

your final integral looked good but the one b4 doesn't click with me. i also thought \\ starts a new line, it works sometimes, and others it doesn't. the \ doesn't even work great for me either.

 

[Saladsamurai]

I edited to fix..I forgot the tan2x where sec2xtan2x=du/2