[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?

[f1mad]

I would let u= sin2x

du/dx = 2cos2x

du = dx*2cos2x.

[ttoby]

Another way of doing this question is to notice that sin(4x)=sin(2x+2x)=2sin(2x)cos(2 x) 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.

[steve2005]

(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?

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


EDIT

Why the rep?

[Luppy021]

Yeah I get it now cheers all

[Zuzuzu]

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

[the bear]

You could also use integration by parts