# Sorry... it's another P5 question... I have a few!

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Q14 in heinemann review exercise P5

inn(between pi/4 and 0)=2/(3sin2x+4cos2x) dx

the anwer =0.2ln6 but once more i can't get there!

q11

I think that there is a mistake/or i'm a twit!

I(n)=int[between1,0]of x^(n-1)xroot(1-x^2) dx

show that (n+2) I(n)=(n-1) I(n-2)

I think that the answer should be (n+1) at the beginning but anyones welcome to have a go!

inn(between pi/4 and 0)=2/(3sin2x+4cos2x) dx

the anwer =0.2ln6 but once more i can't get there!

q11

I think that there is a mistake/or i'm a twit!

I(n)=int[between1,0]of x^(n-1)xroot(1-x^2) dx

show that (n+2) I(n)=(n-1) I(n-2)

I think that the answer should be (n+1) at the beginning but anyones welcome to have a go!

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#2

For Q.14 you can get the answer with a t=tan(x) substitution. If you are not familiar with how to apply these feel free to write back and I will explain, also page 43 example 15 should help.

Does anybody know of a better method for Q.14?

Does anybody know of a better method for Q.14?

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#3

(Original post by

For Q.14 you can get the answer with a t=tan(x) substitution. If you are not familiar with how to apply these feel free to write back and I will explain, also page 43 example 15 should help.

Does anybody know of a better method for Q.14?

**mikesgt2**)For Q.14 you can get the answer with a t=tan(x) substitution. If you are not familiar with how to apply these feel free to write back and I will explain, also page 43 example 15 should help.

Does anybody know of a better method for Q.14?

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#4

(Original post by

u could use 3sin2x+4cos2x=Rsin(2x+a).But i agree with using the t=tanx substitution.

**IntegralAnomaly**)u could use 3sin2x+4cos2x=Rsin(2x+a).But i agree with using the t=tanx substitution.

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#5

(Original post by

I think that the answer should be (n+1) at the beginning but anyones welcome to have a go!

**ogs**)I think that the answer should be (n+1) at the beginning but anyones welcome to have a go!

u = x^(n-1), du/dx = (n-1)x^(n-2)

dv/dx = x(1-x^2)^1/2, v = -(1/3)(1-x^2)^3/2

Applying parts,

I(n) = (n-1)/3 INT x^(n-2) (1-x^2)^3/2 dx

3I(n) = (n-1) INT x^(n-2) (1-x^2)(1-x^2)^1/2 dx

3I(n) = (n-1) INT [ x^(n-2) - x^n ](1-x^2)^1/2 dx

3I(n) = (n-1)[ I(n-2) - I(n) ]

So,

3I(n) + (n-1)I(n) = (n-1)I(n-2)

(n+2)I(n) = (n-1)I(n-2)

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#6

I just used the Rsin(2x+ø) method, but it's a bit tedious - which may explain why I got 0.5ln6 rather than 0.2ln6 !

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(Original post by

I get (n+2) on this question:

u = x^(n-1), du/dx = (n-1)x^(n-2)

dv/dx = x(1-x^2)^1/2, v = -(1/3)(1-x^2)^3/2

Applying parts,

I(n) = (n-1)/3 INT x^(n-2) (1-x^2)^3/2 dx

3I(n) = (n-1) INT x^(n-2) (1-x^2)(1-x^2)^1/2 dx

3I(n) = (n-1) INT [ x^(n-2) - x^n ](1-x^2)^1/2 dx

3I(n) = (n-1)[ I(n-2) - I(n) ]

So,

3I(n) + (n-1)I(n) = (n-1)I(n-2)

(n+2)I(n) = (n-1)I(n-2)

**mikesgt2**)I get (n+2) on this question:

u = x^(n-1), du/dx = (n-1)x^(n-2)

dv/dx = x(1-x^2)^1/2, v = -(1/3)(1-x^2)^3/2

Applying parts,

I(n) = (n-1)/3 INT x^(n-2) (1-x^2)^3/2 dx

3I(n) = (n-1) INT x^(n-2) (1-x^2)(1-x^2)^1/2 dx

3I(n) = (n-1) INT [ x^(n-2) - x^n ](1-x^2)^1/2 dx

3I(n) = (n-1)[ I(n-2) - I(n) ]

So,

3I(n) + (n-1)I(n) = (n-1)I(n-2)

(n+2)I(n) = (n-1)I(n-2)

thanks... I made a ridiculous mistake on the inital integration

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