Integration of trig functions (1 Viewer)

[lacklustre]

Alright guys, I'm in need of a little help with this one:

Integration of: sin³(2x)dx

Here's what I've done:

(the above integration) = Int. sin²(2x).sin(2x)dx
= Int. sin2x(1-cos²(2x))
= Int. sin2x - sin2x.cos²2x



I've not a clue where to proceed from there.

(p.s. how do you do the intergration sign)

 

[Slidey]

sin2x is the derivative of cos2x, so use chain rule. Or the reverse of it (substitution).

use: u=cos(2x)
du=-2sin(2x) dx

 

[Mark576]

∫sin³(2x)dx
Let u=cos(2x)=>du/dx=-2sin(2x)=>dx=-du/2sin(2x) [using Slidey's substitution]
I = -1/2∫sin²(2x)du
Using sin²(2x) = 1 - cos²(2x);
I = -1/2∫1-cos²(2x)du = -1/2∫1-u²du = -1/2(u - u³/3) = -1/2[cos(2x) - cos³(2x)/3] + c

 

[lacklustre]

ah thank you kind sirs.

 

[lacklustre]

I'm in another sticky situation with:

Int. sin⁴x dx

Do I sub in sin²x = (1-cos2x)/2 ?

 

[cwag]

yea...you could do that....

int: sin²xsin²x
= 1/4(1-cos 2x)²
=1/4(1-2cos2x + 1/2(1+cos4x))
=1/4(1- 2cos 2x + 1/2 + (1/2)cos4x
=3/8 - (1/2)cos2x + (1/8)cos4x

integral becomes,
3x/8 - (1/4)sin 2x + (1/32)sin 4x + C

 

[lacklustre]

Thanks cwag.

yea...you could do that....

Is there another way? and is it faster?

 

[cwag]

Is there another way? and is it faster?

hmm...i dunno...that seems to me like the easiest way...sorry if i sounded derogatory...i was just thinking

 

[lacklustre]

Stuck on a few more: (but most of the other types I am great with)

a) Int. (between 0 ---> 1/2) sqrt.(1-x²) dx, by letting x = sinө

b) Int. (between 0 ---> 1) sqrt.(x/(1-x)) dx, letting x = sin²ө

cheers

 

[lacklustre]

If you don't wish to do the whole question could anybody at least tell me where to go?

 

[pLuvia]

a. Let x=sin@
dx=cos@ d@
Then when x=0, @=0
when x=1/2, @=pi/6

int. sqrt(1-sin²@) cos@ d@
=int. sqrt(cos²@) cos@ d@
=int. cos²@ d@
=int. (cos2@+1)/2 d@
=1/2[(sin2@)/2+@]
Shove in the limits then you get your answer

b. Let x=sin²@
dx=2sin@cos@ d@
When x=0, @=0
when x=1, @=pi/2

int. sqrt[(sin²@)/(1-sin²@)] 2sin@cos@ d@
=int. sin@/cos@* 2sin@cos@ d@
=int. 2sin²@ d@
=int. (cos2@-1)/2 d@

then do the same and you get your answer

 

[lacklustre]

^^ ahh cheers. Thank you very much.