Calculus

[Merkwürdigeliebe]

I need some help, I just can't figure this out...

What method do I use? I still haven't figured out integration by parts and substitution is a bit tricky for me too. Can someone do this for me step by step? I'd appreciate it so much.


edit:

bah nevermind, I figured it out eventually, integration by parts:


let u=x, dv=(tan^2)x

~x(tan^2)xdv = ~udv

= uv - ~vdu

= x(tan^2)x - ~(tan^2)xdx

= x(tan^2)x + 2ln|cosx| + C


I think....

no wait, my bad, I think it's actually -2xln|cosx| + 2ln|cosx| + C


ah well the ****, it's all wrong

 

[Merkwürdigeliebe]

meh, I figured it out if anyone's even remotely interested :freak:

argh, I always forget to put "+ C"

 

[peterhuang913]

wait... if u=x, how did you get to dv=tan(x)^2

Can you explain how you did it. Maybe do it again in one column?

Do I read it first column then second column?

 

[Merkwürdigeliebe]

I hope this explains it better:

uv minus integral of (v)du is the formula you use in integration by parts

 

[bobojuice]

im so glad i never took any math after high school.

 

[Ronnie]

^lol i never took math after my sophmore year in high school

 

[peterhuang913]

 

[Merkwürdigeliebe]

The methods used are 'Integration by Parts' which also uses 'Integration by Substitution'

You break the expression into two pieces - in this case 'x' and 'tan²x', and then you assign each to either 'u' or 'dv'

in this case, u=x and dv=tan²x

I did in fact miss an integral sign on the second line, my bad..

dx is gone because it is in terms of dv, by the third line, it's in terms of du, and then when you expand u and v it becomes in terms of dx

the third line is a set formula you use for 'integration by parts'

 

[peterhuang913]

So shouldn't the second line be / udv dx?

 

[Merkwürdigeliebe]

So shouldn't the second line be / udv dx?

no because it's in terms of dv