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    Can someone please show how to do these type of questions?:mad:

    Intergrate cos^4(x)sin^3(x) dx
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    (Original post by Dopey')
    Can someone please show how to do these type of questions?:mad:

    Intergrate cos^4(x)sin^3(x) dx
    Well with that one you use sin^2x = 1 - cos^2x

    Then you have 2 parts that look like cos^nxsinx so you use inverse chain rule
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    (Original post by TenOfThem)
    Well with that one you use sin^2x = 1 - cos^2x

    Then you have 2 parts that look like cos^nxsinx so you use inverse chain rule
    cos^4(x)sinx(sin^2(x))
    cos^4(x)sinx(1-cos(x))]

    By the way its not cos to the power of 4x its cos to the power of 4
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    (Original post by Dopey')
    cos^4(x)sinx(sin^2(x))
    cos^4(x)sinx(1-cos(x))]

    By the way its not cos to the power of 4x its cos to the power of 4
    Yeah I know

    Your working above is incorrect sin^2x=1-cos^x

    Then multiply the bracket
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    (Original post by TenOfThem)
    Yeah I know

    Your working above is incorrect sin^2x=1-cos^x

    Then multiply the bracket
    cos^4(x)sinx(1-cos^2x)
    cos^4(x)sinx -cos^6(x)sinx

    Its this ok so far, what the next step?

    think ive got it

    i did

    cos^4(x)sinx(1-cos^2(x)) then u = cos x u' = -sinx dx/du = -cosec x

    put u in

    - u^4 + u^6 then it grate and substitute cos x back in after
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    (Original post by Dopey')
    cos^4(x)sinx(1-cos^2x)
    cos^4(x)sinx -cos^6(x)sinx

    Its this ok so far, what the next step?

    think ive got it

    i did

    cos^4(x)sinx(1-cos^2(x)) then u = cos x u' = -sinx dx/du = -cosec x

    put u in

    - u^4 + u^6 then it grate and substitute cos x back in after
    Will the same technique apply to intergrating sin^3x dx ?

    my steps so far

    Sinx(sin^2x)
    Sinx(1-cos^2x) Now I cant see what to do?
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    (Original post by Dopey')
    Will the same technique apply to intergrating sin^3x dx ?

    my steps so far

    Sinx(sin^2x)
    Sinx(1-cos^2x) Now I cant see what to do?
    Expand the bracket. Now integrate term by term (you can use a substitution for the second term).
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    (Original post by Mr M)
    Expand the bracket. Now integrate term by term (you can use a substitution for the second term).
    I think im getting the hang of it, but it always seems to get harder could you show how you would solve this please

    Intergrate in(rt(x-1))dx
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    (Original post by Dopey')
    I think im getting the hang of it, but it always seems to get harder could you show how you would solve this please

    Intergrate in(rt(x-1))dx
    Why does everyone think the natural logarithm is written as "in"?

    Make a substitution u=x-1

    Remember than \ln(x^n)=n \ln x

    Integrate by parts.
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    (Original post by Mr M)
    Why does everyone think the natural logarithm is written as "in"?

    Make a substitution u=x-1

    Remember than \ln(x^n)=n \ln x

    Integrate by parts.
    Haha I dunno why everyone does, It just one of those things i guess

    One more question I'm attempting to intergrate x^5e^(x^3)

    I intergrate by parts but it just goes round in circle (x keeps coming back) is there a special technique for these type of questions?
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    (Original post by Dopey')
    Haha I dunno why everyone does, It just one of those things i guess

    One more question I'm attempting to intergrate x^5e^(x^3)

    I intergrate by parts but it just goes round in circle (x keeps coming back) is there a special technique for these type of questions?
    You just need lots of experience.

    Write x^5 as x^3 x^2 and make a substitution u=x^3.
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    (Original post by Mr M)
    You just need lots of experience.

    Write x^5 as x^3 x^2 and make a substitution u=x^3.
    so is dv/dx = x^2e^(x^3)?

    v = 1/3e^(x^3)?

    I think Ive done it the answers right but the workings looks horrible, maybe theres a quicker method?

    u = x^3

    u' = 3x^2

    v = 1/3u v' = 1/3
    dz/dx = e^u z= e^u

    = 1/3ue^u - (intergrate) 1/3e^u
    = 1/3ue^u - 1/3e^u
    = 1/3x^3e(x^3) - 1/3e^(x^3) = 1/3e^(x^3)(x^3-1) + c

    Is this ok?
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    (Original post by Dopey')
    so is dv/dx = x^2e^(x^3)?

    v = 1/3e^(x^3)?
    You seem to have failed to make the substitution I suggested?
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    (Original post by Dopey')
    so is dv/dx = x^2e^(x^3)?

    v = 1/3e^(x^3)?
    That'll work.
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    (Original post by Mr M)
    You seem to have failed to make the substitution I suggested?
    Ive a adjusted my workings above ^ (how you suggested)
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    (Original post by Dopey')
    the workings looks horrible, maybe theres a quicker method?
    Maybe MAKING A SUBSTITUTION would help?



    \displaystyle \int x^5 e^{x^3} \, dx = \frac{1}{3} \int t e^t \, dt
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    (Original post by Mr M)
    Maybe MAKING A SUBSTITUTION would help?



    \displaystyle \int x^5 e^{x^3} \, dx = \frac{1}{3} \int t e^t \, dt
    I did substitute u = x^3 in, If you didnt mean it like this i dont understand im a novice
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    (Original post by Dopey')
    I did substitute u = x^3 in, If you didnt mean it like this i dont understand im a novice
    My fault I think. The choice of u was confusing for you as you muddled it with u from integration by parts. This is why I changed it to t above.
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    (Original post by Mr M)
    Maybe MAKING A SUBSTITUTION would help?



    \displaystyle \int x^5 e^{x^3} \, dx = \frac{1}{3} \int t e^t \, dt
    To be fair to Dopey, he saw your breakdown of x^5 with u=x^3 and thought you were doing IBP with dv/dx equal the rest.
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    (Original post by notnek)
    To be fair to Dopey, he saw your breakdown of x^5 with u=x^3 and thought you were doing IBP with dv/dx equal the rest.
    Yes I realise that. I failed to communicate properly.
 
 
 
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