Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

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Integration (C4) 4sin2xcos2x

Integration of 4sin2x.cos2x dx

Not sure how to do this but I tried by substitution and got -cos^2(2x)+c

u=cos2x
du/dx=-2sin2x
dx/du=-1/2sin2x
sin2x(dx/du)=-1/2
But arrived at the wrong answer is there a quicker qay of doing this?

I would let u= sin2x

du/dx = 2cos2x

du = dx*2cos2x.

Another way of doing this question is to notice that sin(4x)=sin(2x+2x)=2sin(2x)cos(2x) then using that to make a trig substitution.

However, the answer you got is actually correct. If you try my method you get (-1/2)cos(4x) + k which can be manipulated to show that (-1/2)cos(4x) differs from -cos^2(2x) by a constant and hence they are equivilant answers to the question.

Original post by Luppy021

Integration of 4sin2x.cos2x dx

Not sure how to do this but I tried by substitution and got -cos^2(2x)+c

u=cos2x
du/dx=-2sin2x
dx/du=-1/2sin2x
sin2x(dx/du)=-1/2
But arrived at the wrong answer is there a quicker qay of doing this?

Uploaded with ImageShack.us

There are several ways of doing this problem and they might get answers that LOOK different.

EDIT

Why the rep?

(edited 12 years ago)

OP

Yeah I get it now cheers all

Your answer is correct. I think it will differ by a constant from the book answer.

You could also use integration by parts

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