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# Integrate Sin^3(2x) - Help Please watch

1. Put me out of my misery please!
2. Write and then use the identity . You can then integrate the resulting two terms separately; if you get stuck on this let us know.
3. So do I end up with

Integral of Sin2x multiplied by the integral of (1-cos^(2x))??

Even so I dont' understand why I would need to convert sin^2x to (1-cos^2x) sorry if I'm missing something because this it what came up when I searched on Google but I don't understand.

You can multiply integrals together can't you?
4. Ignore that, I get it!!!!

I think anyway, let me have a crack at a few in the book and I'll come back if I get stuck. Thanks for your help and time
5. Actually, I've got this far: -

Int (sin2x) - Int (Cos^2x)(Sin2x) <-- How to I integrate this end bit?

I'm fine with the first bit (sin2x)
6. (Original post by rosschambers1987)
Actually, I've got this far: -

Int (sin2x) - Int (Cos^2x)(Sin2x) <-- How to I integrate this end bit?

I'm fine with the first bit (sin2x)
What do you get when you differentiate ?
7. I don't know, my brain isn't working (bangs head on desk).

Is it just chain rule and I get

3(cos2x)^2 * 1/2(sin2x)? so it's 3/2(cos^2x)(sin2x) so it would be 2/3(cos^3(2x))?

I'm nearly there, the answer in the book has a +1/6 instead of my -2/3.
8. Sorry I've got it now, started from scratch again, got the same answer as in the book (I didn't realise I had to chain rule twice on differtiating cos^3(2x).

Thanks so much.
9. I'm not sure if this is right but I gave it a go...

Integral of sin^3(2x) dx

well first you can factorise sin(2x) to get a positive power --->> the integral of sin(x)[(sin^2(2x)]

then use the trig. identity sin^2(x)+cos^2(x)=1 so the integral of (1-cos^2(2x))sin(x) dx

now use substitution, u=cos(2x) the derivative of that is -2sin(2x)

there dx=du/-2sin(2x)

so the integral of (1-u^2)sin(2x) du/-2sin(2x)

you can cancel out the sin(2x) function and factorise out the negative half to get

-1/2 integral 1-u^2 du

then just integrate to get -1/2(u-u^3/3)+c

then you get -1/2cos(2x)- 1/3cos^3(2x)+c

I may be wrong so check out this video on youtube and see if you can follow the steps
10. (Original post by A15hat)
I'm not sure if this is right but I gave it a go...

Integral of sin^3(2x) dx

well first you can factorise sin(2x) to get a positive power --->> the integral of sin(x)[(sin^2(2x)]

then use the trig. identity sin^2(x)+cos^2(x)=1 so the integral of (1-cos^2(2x))sin(x) dx

now use substitution, u=cos(2x) the derivative of that is -2sin(2x)

there dx=du/-2sin(2x)

so the integral of (1-u^2)sin(2x) du/-2sin(2x)

you can cancel out the sin(2x) function and factorise out the negative half to get

-1/2 integral 1-u^2 du

then just integrate to get -1/2(u-u^3/3)+c

then you get -1/2cos(2x)- 1/3cos^3(2x)+c

I may be wrong so check out this video on youtube and see if you can follow the steps
It is almost right.
At the last row the factor of cosine function will be (-1/2)(-1/3)=1/6

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