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    So I've had a sheet of 50 integration questions to do and several on the sheet have so far stumped me.

    Any help would be appreciated thanks!

    9) Sinxcos^(4)x

    13) Sin^(2)2x

    18) x/9x^2 + 1

    24) cot^(2)3x

    31) (x+1)^2/x^2+1.
    I used division to work out it was the integral of 1 +(2x/x^2+1) Then how do i progress further?
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    (Original post by Henry.Lister)
    So I've had a sheet of 50 integration questions to do and several on the sheet have so far stumped me.

    Any help would be appreciated thanks!

    9) Sinxcos^(4)x

    13) Sin^(2)2x

    18) x/9x^2 + 1

    24) cot^(2)3x

    31) (x+1)^2/x^2+1.
    I used division to work out it was the integral of 1 +(2x/x^2+1) Then how do i progress further?
    For many of them, you have to consider the chain rule in reverse (when you have a function multiplied by it's differential). Others involve natural logs. Does this help?
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    Hmm i recognise where you're coming from but I'm weary about knowing exactly how to do it.

    Mind showing me one? Thanks
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    (Original post by Henry.Lister)
    Hmm i recognise where you're coming from but I'm weary about knowing exactly how to do it.

    Mind showing me one? Thanks
    For 9), consider the differential of cos^(5)x.

    For 18), consider the differential of ln(9x^2 +1)
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    (Original post by Henry.Lister)
    So I've had a sheet of 50 integration questions to do and several on the sheet have so far stumped me.

    Any help would be appreciated thanks!

    9) Sinxcos^(4)x

    13) Sin^(2)2x

    18) x/9x^2 + 1

    24) cot^(2)3x

    31) (x+1)^2/x^2+1.
    I used division to work out it was the integral of 1 +(2x/x^2+1) Then how do i progress further?
    For 9)
    Use u=cosx substitution
    du/dx=-sinx -> -sinx dx =du
    or use directly the rule of
    \int f'(x)\cdot f^n(x) dx =\frac{f^{n+1}(x)}{n+1}+C
    that is
    \int sinx \cdot cos^4xdx=-\int (-\sin x)\cdot (\cos x)^4 dx

    for 13)
    Use the identity of
    \sin^2 A=\frac{1-\cos 2A}{2}
    and
    when the primitive function for f(x) is F(x)+C
    then the primitive function for f(ax+b) is F(ax+b)/a +C

    for 18)
    \int \frac{f'(x)}{f(x)} dx =\ln |f(x)| +C
    For this quesion
    \int \frac{x}{9x^2+1} dx=\frac{1}{18}\int \frac{18x}{9x^2+1} dx
    But you can use substitution of t=9x^2+1
    then dt=18x dx -> x dx =1/18 dt
    \int \frac{x}{9x^2+1} dx=\int \frac{1}{9x^2+1} \cdot x \cdot dx=\int \frac{1}{t}\cdot \frac{1}{18} dt

    for 24)
    cot^2 3x=\frac{\cos^2 3x}{\sin^2 3x}=\frac{1-sin^2 3x}{sin^2 3x}=\frac{1}{sin^2 3x}-1
    And as you know -\int \frac{1}{\sin^2 (ax+b)} dx =\frac{\cot (ax+b)}{a} +C

    for 31)
    see 18
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    (Original post by ztibor)
    x
    You're not supposed to give solutions on these forums, but to help the OP reach the answer themselves
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    9 makes sense now thanks. I'll look through 18 next.
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    Ztibor thanks, I'll look through those calculations shortly
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    I think I've managed to complete them all except I'm unsure about 13.

    I tried changing the integral of sin^22x into (1-cos4x)/2 using cos2x=1-2sin^2x

    I then integrated that to get 1/2x-1/4sin4x+c.
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    (Original post by Henry.Lister)
    I think I've managed to complete them all except I'm unsure about 13.

    I tried changing the integral of sin^22x into (1-cos4x)/2 using cos2x=1-2sin^2x

    I then integrated that to get 1/2x-1/4sin4x+c.
    This is the correct method - but you've made a small slip up in the solution.
    Spoiler:
    Show
    Reverse chain rule on -cos4x.
 
 
 
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