# Reduction formulae

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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 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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#2

(Original post by

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?

**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?

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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(Original post by

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.

**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.

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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#4

(Original post by

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..

**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..

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#5

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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#6

(Original post by

golgiapparatus31 type on google

**TeeEm**)golgiapparatus31 type on google

**and look towards the end of the booklet***madas reduction formulas*
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**TeeEm**)

golgiapparatus31 type on google

**and look towards the end of the booklet**

*madas reduction formulas*
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(Original post by

Did you get it sorted?

**mqb2766**)Did you get it sorted?

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