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1. i have a couple of questions im stuck on to do with p3

1) how do you integrate (lny)/y ?

2)
how do you integrate (sin3x - cos3x)/(sin2x cos2x)

this question is the last part of quite a long question, in the first parts i found that;
d(secx)/dx = secx tanx
d(cosecx)/dx = -cosecx cotx
cosecx + secx = 2(sinx + cosx)/sin2x
and d((sinx + cosx)/sin2x)/dx = 0.5(tanx secx - cotx cosecx)
i figure that to answer question 2, these identities must be used somehow, but i'm not quite sure how. its only worth 3 marks so it can't be too complicated.

thanks
2. 1. See yazan's post below. 

2.
∫ (sin³x - cos³x)/(sin²x cos²x) dx = ∫ sin³x/(sin²x cos²x) dx - ∫ cos³x/(sin²x cos²x) dx
= ∫ sinx/cos²x dx - ∫ cosx/sin²x dx
= ∫ tanx secx dx - ∫ cotxcosecx dx
= secx + cosecx + C
3. 1) how do you integrate (lny)/y ?
use u = lny
du/dy = 1/y
dy = y du

int ln(y) / y dy = int u/y .ydu = int u du = ½u² +C = ½(lny)² + C
4. (Original post by gemm)
i have a couple of questions im stuck on to do with p3

1) how do you integrate (lny)/y ?

2)
how do you integrate (sin3x - cos3x)/(sin2x cos2x)

this question is the last part of quite a long question, in the first parts i found that;
d(secx)/dx = secx tanx
d(cosecx)/dx = -cosecx cotx
cosecx + secx = 2(sinx + cosx)/sin2x
and d((sinx + cosx)/sin2x)/dx = 0.5(tanx secx - cotx cosecx)
i figure that to answer question 2, these identities must be used somehow, but i'm not quite sure how. its only worth 3 marks so it can't be too complicated.

thanks
for part 2:

(sin3x - cos3x)/(sin2x cos2x)

this is equal to: (sin3x)/(sin2x cos2x) - (cos3x)/(sin2x cos2x)

this simplifies to sinx/cos2x - cosx/sin2x
which is secx tanx - cosecx cotx

INT (secx tanx - cosecx cotx) dx
= secx + cosecx + C
5. cheers,

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