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# Integrating cot^2 2x dx. Rep for help. watch

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1. Hi
Can anyone integrate cot^2 2x and show me how you did it please? The answer in the book is -x - 1/2 cot x + c.

Rosie <== crap at p3
2. (Original post by crana)
Hi
Can anyone integrate cot^2 2x and show me how you did it please? The answer in the book is -x - 1/2 cot x + c.

Rosie <== crap at p3
Hang on a sec... I've got the answer.
3. Dividing

sin^2(2x) + cos^2(2x) = 1

by sin^2(2x), we get

1 + cot^2(2x) = cosec^2(2x).

So we want to integrate -1 + cosec^2(2x). Most A-level formula booklets (eg, http://www.aqa.org.uk/admin/library/gce-maths-bklet.pdf) tell you that the derivative of cot(x) is -cosec^2(x). By the chain rule it follows that the derivative of cot(2x) is -2 cosec^2(2x). So the integral of cosec^2(2x) is -(1/2) cot(2x) + constant.

(int) cot^2(2x) dx
= (int) -1 + cosec^2(2x) dx
= -x - (1/2) cot(2x) + constant.
4. (Original post by crana)
Hi
Can anyone integrate cot^2 2x and show me how you did it please? The answer in the book is -x - 1/2 cot x + c.

Rosie <== crap at p3
∫cot²(2x) dx

= ∫(cos²(2x))/(sin²(2x)) dx

= ∫cos(2x) * (cos(2x))/(sin²(2x)) dx

= 1/2 * ∫cos(2x) * (sin(2x))' * (sin(2x))^(-2) dx

=...
5. ∫cot^2(2x) dx

let u = 2x

∫cot^2(u) dx

By definition:

cot^2(u) = cosec^2(u) -1

so:

∫cosec^2(u) -1 dx

du/dx = 2

so dx = du/2

(1/2)∫cosec^2(u) -1 du

given in the formula book:

"derivative of cot(@) is -cosec^2(@)"
(ie cosec^2(@) integrates to -cot(@) )

so integral goes to:

(1/2).(-cot(u)) - u + C

= -(1/2)cot(2x) - x + C

Hope this helps
6. (Original post by zizero)
∫cot²(2x) dx

= ∫(cos²(2x))/(sin²(2x)) dx

= ∫cos(2x) * (cos(2x))/(sin²(2x)) dx

= 1/2 * ∫cos(2x) * (sin(2x))' * (sin(2x))^(-2) dx

=...
f(x) = cos(2x) ; g'(x) = (sin(2x))' * (sin(2x))^(-2)
f'(x) = 2sin(2x) ; g(x) = -(sin(2x))^(-1) = -1/(sin(2x))

...
7. (Original post by zizero)
f(x) = cos(2x) ; g'(x) = (sin(2x))' * (sin(2x))^(-2)
f'(x) = 2sin(2x) ; g(x) = -(sin(2x))^(-1) = -1/(sin(2x))

...
So, we get: -0,5*(cos(2x))/(sin2x) - 0,5*2*∫sin(2x)/sin(2x) dx

= -1/2 * cot2x - x + k
8. Cheers everyone

ill distribute some rep as soon as I can again
rosie

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