golgiapparatus31
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#1
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#1
https://imgur.com/a/J9aZ6h7

For 6,
u = (1-2x^4)^n dv = x dx

so I = (-1)^n/2 + integral of (4nx^5 ( 1-2x^4)^n-1) dx

Not sure what to do now.. tried to use u = x^2 and dv = (-8x^3)(1-2x^4)^n-1 but no progress

For 3,

used u = x^4 and dv = x^2(1+x^3)^4 dx

I4 = 32/15 - 4/15 integral of x^3(1+x^3)^5 dx

now used (1+x^3)^5 = (1+x^3)(1+x^3)^4

got 19/15 I_4 = 32/15 - 4/15 integral of (x^3(1+x^3)^3)
Stuck again...
answer given for this part is I_4 = 16/19 + 12/19I_3

I didnt get there. Tried a variety of stuff, reached I4 = I3 + 11281/19760

Can someone please give me some pointers on how to proceed with this?
Last edited by golgiapparatus31; 4 weeks ago
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mqb2766
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#2
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#2
(Original post by golgiapparatus31)
https://imgur.com/a/J9aZ6h7

For 6,
u = (1-2x^4)^n dv = x dx

so I = (-1)^n/2 + integral of (4nx^5 ( 1-2x^4)^n-1) dx

Not sure what to do now.. tried to use u = x^2 and dv = (-8x^3)(1-2x^4)^n-1 but no progress

For 3,

used u = x^4 and dv = x^2(1+x^3)^4 dx

I4 = 32/15 - 4/15 integral of x^3(1+x^3)^5 dx

now used (1+x^3)^5 = (1+x^3)(1+x^3)^4

got 19/15 I_4 = 32/15 - 4/15 integral of (x^3(1+x^3)^3)
Stuck again...
answer given for this part is I_4 = 16/19 + 12/19I_3

I didnt get there. Tried a variety of stuff, reached I4 = I3 + 11281/19760

Can someone please give me some pointers on how to proceed with this?
For the first one you're 1/2 way there. REwrite the original problem as
I_n = Int x(1-2x^4)^n = Int x(1-2x^4)^(n-1)(1-2x^4)
= I_(n-1) - Int 2x^5(1-2x^4)^(n-1)
The second term here is 2n out from your extra integral, so replace and rearrange the expression.
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golgiapparatus31
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#3
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#3
(Original post by mqb2766)
For the first one you're 1/2 way there. REwrite the original problem as
I_n = Int x(1-2x^4)^n = Int x(1-2x^4)^(n-1)(1-2x^4)
= I_(n-1) - Int 2x^5(1-2x^4)^(n-1)
The second term here is 2n out from your extra integral, so replace and rearrange the expression.
Thanks so much!! I managed to reach the answer!

For 3, I tried again, continuing where I was stuck in post 1,
I used (1+x^3)^4 = (1+x^3)(1+x^3)^3
to rewrite the integral,

getting 19/15 I_4 = 32/15 - 4/15 I_3 - 4/15 integral of (x^3(1+x^3)^3) dx
I'm stuck so I just evaluate the integral, by expanding it and integrating term by term, getting

19/15 I_4 = 32/15 - 4/15 I_3 - 1921/6825
which does not simplify to the given form..
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mqb2766
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#4
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#4
(Original post by golgiapparatus31)
Thanks so much!! I managed to reach the answer!

For 3, I tried again, continuing where I was stuck in post 1,
I used (1+x^3)^4 = (1+x^3)(1+x^3)^3
to rewrite the integral,

getting 19/15 I_4 = 32/15 - 4/15 I_3 - 4/15 integral of (x^3(1+x^3)^3) dx
I'm stuck so I just evaluate the integral, by expanding it and integrating term by term, getting

19/15 I_4 = 32/15 - 4/15 I_3 - 1921/6825
which does not simplify to the given form..
Will look at the other one in the morning. Too tired now :-(
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DFranklin
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For Q3, it looks like it would be easier to start with I3 and then integrating by parts will give you something that looks like I4.
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Prince Philip
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#6
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#6
(Original post by TeeEm)
golgiapparatus31 type on google madas reduction formulas and look towards the end of the booklet
This Madas guy seems to know his stuff.
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golgiapparatus31
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#7
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(Original post by TeeEm)
golgiapparatus31 type on google madas reduction formulas and look towards the end of the booklet
Thanks!!!
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mqb2766
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(Original post by golgiapparatus31)
Thanks!!!
Did you get it sorted?
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golgiapparatus31
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#9
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(Original post by mqb2766)
Did you get it sorted?
Yes, I managed to solve the problem. Thank you for the help!!
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