integration

manps

Hi, i need help with these please

Integrate:

cot^2 (3x)

and

x/(x-1)^½

Thanks

Reply 1

username9816

14

manps

Integrate:
cot^2 (3x)
and
x/(x-1)^½

Int. cot^2(3x) dx. = Int. cosec^2(3x) - 1 dx. = [-cot(3x)]/3 - x + k = k - [x + cot(3x)/3]

Int. x/[(x - 1)^(1/2)] dx.

Let u = x - 1 -> du/dx = 1 -> dx = du

Hence: Int. x/[(x - 1)^(1/2)] dx. = Int. (u + 1)/[u^(1/2)] du = Int. u^(1/2) + u^(-1/2) du. = [2u^(3/2)]/3 + 2u^(1/2) + k = [2(x - 1)^(3/2)]/3 + 2(x - 1)^(1/2) + k

Reply 2

manps

OP

Thanks a lot for your help

ALso, i have seen a lot of integration questions like this one.

INT sin2xcosx

The only way i can think of doing these is INT by parts, getting 2 equations and subtracting out art of it, but im not too sure on how to do this method.

Is there a more efficient way of doing ones like these?

Another one is INT cosxcos2x

Reply 3

Gaz031

15

manps

Thanks a lot for your help

ALso, i have seen a lot of integration questions like this one.

INT sin2xcosx

The only way i can think of doing these is INT by parts, getting 2 equations and subtracting out art of it, but im not too sure on how to do this method.

Is there a more efficient way of doing ones like these?

Another one is INT cosxcos2x

INT sin2xcosx dx = INT 2sinx(cosx)^2 dx
Use u=cosx.

INT cosxcos2x dx = INT cosx(1-2sin^2x) dx = INT cosx dx - INT 2cosx(sinx)^2 dx
The first part is a standard result but for the second part use u=sinx.

integration

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