A bit like the cos() and sin() pairing (derivative and pythagorean identity), tan() and sec^2() have a similar relationship so cot() and .... So think how the trig terms are related, then when you do the parts, you want to differentate the x^2 and integrate the trig term(s)?
(edited 2 years ago)
Original post by mqb2766
A bit like the cos() and sin() pairing (derivative and pythagorean identity), tan() and sec^2() have a similar relationship so cot() and .... So think how the trig terms are related, then when you do the parts, you want to differentate the x^2 and integrate the trig term(s)?
Oh so you mean 1+cot²=cosec² , so the trig parts simplify to cot³x- cotx , so I can split it into 2 integrals?
And yes I'm differentiating the x² and integrating the trig
If it is the case, I've gone terribly wrong as I'm now stuck in an infinite integration loop with the 2nd integral :/
(edited 2 years ago)
Original post by mqb2766
No, use the derivative relatisnhip s0
d cot(x)/dx = - cosec^2(x)
d cot(x)/dx = - cosec^2(x)
Use it how?
LIATE so u=x² and dv/dx = cosec²xcotx
Du/dx = 2x And to get v do we then integrate by reverse chain rule as it's f'(x)f(x) ?
(edited 2 years ago)
Original post by mqb2766
You want to integrate
f'(x)f(x)
So the integral should be cosec^2().
f'(x)f(x)
So the integral should be cosec^2().
Meaning dv/dx = cosec²x and u is 2x²cotx ?
Original post by Rhys_M
Meaning dv/dx = cosec²x and u is 2x²cotx ?
no you were right the first time.
If you're stuck on how to integrate cotxcosec^2x, consider the substitution u=cotx.
(edited 2 years ago)
Original post by JustSomeGuy:/
no you were right the first time.
If you're stuck on how to integrate cotxcosec^2x, consider the substitution u=cotx.
If you're stuck on how to integrate cotxcosec^2x, consider the substitution u=cotx.
Yea That's what I thought - I was then doing that subsitution. Thank you
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