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    Would much appreciate some help on this question:

    Integrate: {sin}^2{x}    {cos}^2{x}

    Using trig functions/identities involving cos 2x, I got:

    \int (\frac{1}{2} - \frac{1}{2}cos 2x)(\frac{1}{2}cos 2x + \frac{1}{2})\ dx









= \int \frac{1}{4} - \frac{1}{4}{cos}^2{2x}\ dx









= \int \frac{1}{4} - \frac{1}{4}(\frac{1}{2}cos 2x + \frac{1}{2})\ dx









= \int \frac{1}{4} - \frac{1}{8} cos 2x - \frac{1}{8}\ dx = \frac{1}{8}x - \frac{1}{16}sin 2x + c

    ......but they got  - \frac{1}{32} sin2x instead of  - \frac{1}{16} sin 2x for the final answer?

    I think I might have gone wrong on the third line with the substitution of  {cos}^2{2x}?

    Also, how do you integrate  \frac{1}{{sin}^2{x}   {cos}^2{x}}? :confused:
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    sinAcosA = \frac{1}{2}sin2x

    This should help.

    EDIT: What the above post said basically ^^
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    (Original post by soutioirsim)
    sinAcosA = \frac{1}{2}sin2x

    This should help.

    EDIT: What the above post said basically ^^
    For integrating the reciprocal, you mean?
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    I suck with Latex but I'll give it a go

    I got x/8 - Sin4x / 32 +C :eek:

    I used (1/2 sin2x)^2

    Expanded then found 1/4 Sin^2 2x in terms of cos4x.

    I think I may have made a mistake aswell somewhere is something wrong with the 1/4 Sin^2 2x into terms of cos4x ?
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    (Original post by Introverted moron)
    For integrating the reciprocal, you mean?
    Yeah, so you have:

    \frac{1}{(\frac{1}{2}sin2x)^2}

    Can you see what to do from there?
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    (Original post by soutioirsim)
    Yeah, so you have:

    \frac{1}{(\frac{1}{2}sin2x)^2}

    Can you see what to do from there?
    Convert that to }{(\frac{1}{2}sin2x)^{-2} and use the chain rule? :erm:
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    i found when you got to integral 1/4 - 1/4(cos^2 2x) dx...

    when you replace the (cos^2 2x) with (1/2cos2x + 1/2) it should be (1/2cos4x +1/2) so then when you divide by differential should give 1/32?
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    sorry didnt really explain. you replaced it with the double angle identity as if it was just (cos^2 x) when in this case it was (cos^2 2x) so you had to multiply the x by 2.
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    (Original post by Introverted moron)
    Convert that to }{(\frac{1}{2}sin2x)^{-2} and use the chain rule? :erm:
    You can't use the chain rule because this isn't differentiation. If you were referring to the reverse chain rule, you can't use that either because the integrand isn't of the required form. Instead notice that \dfrac{1}{\sin ^2 (2x)} \equiv cosec ^2 (2x), which is a standard integral.
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    just gone through it again and not totally sure thats right, because that would give the final answer 1/8x - 1/32 sin4x +c which doesn't sound right....
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    (Original post by gbsn350)
    sorry didnt really explain. you replaced it with the double angle identity as if it was just (cos^2 x) when in this case it was (cos^2 2x) so you had to multiply the x by 2.
    So if  cos 2x = 2{cos}^{2}x - 1, then  2{cos}^{2}2x - 1 must equal  cos 4x?
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    (Original post by Introverted moron)
    Convert that to }{(\frac{1}{2}sin2x)^{-2} and use the chain rule? :erm:
    Your book has the answer wrong I just checked with wolfram alpha and the actual answer is:

    x/8 - 1/32 sin4x +C

    (for the first question)

    For the second, I would write 1/ (1/4 Sin^2 2x) as 4cosec^2 2x

    Then you can integrate 4 cosec^2 2x to become -2cot2x using the formula in your formula booklets.

    Integral Cosec^2 x = -cotx

    Hope this helps Wolfram alpha is really useful to check your answers with

    http://www.wolframalpha.com/input/?i...x%29+integrate
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    (Original post by gbsn350)
    just gone through it again and not totally sure thats right, because that would give the final answer 1/8x - 1/32 sin4x +c which doesn't sound right....
    Thats the correct answer the book must be wrong, just checked it up on wolfram alpha
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    (Original post by Farhan.Hanif93)
    You can't use the chain rule because this isn't differentiation. If you were referring to the reverse chain rule, you can't use that either because the integrand isn't of the required form. Instead notice that \dfrac{1}{\sin ^2 (2x)} \equiv cosec ^2 (2x), which is a standard integral.
    Ah, I see. I'll have a go at it later on/tomorrow using your pointers; I know it's not difficult from here, but my brain's a little fried at the moment.

    Thank you and everyone else who has posted for your help.
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    (Original post by Phil1541)
    Thats the correct answer the book must be wrong, just checked it up on wolfram alpha
    No, the book's right. I'm wrong. It is sin 4x.
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    basically what im saying is... at the beginning you subbed in (1/2 cos2x + 1/2) for cos^2 x.

    then on the third line you've got cos^2 2x but you've subbed in (1/2 cos2x +1/2) again which can't be right because its a 2x not just an x.

    so you should of subbed in (1/2 cos4x + 1/2)
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    (Original post by gbsn350)
    basically what im saying is... at the beginning you subbed in (1/2 cos2x + 1/2) for cos^2 x.

    then on the third line you've got cos^2 2x but you've subbed in (1/2 cos2x +1/2) again which can't be right because its a 2x not just an x.

    so you should of subbed in (1/2 cos4x + 1/2)
    Yeah, I can see the logic behind that now. Thanks for your help.
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    (Original post by Introverted moron)
    Yeah, I can see the logic behind that now. Thanks for your help.
    It is often useful when to integrating (1/something) to bring it onto the top. With trig it will often have a different name. With e^x you will get a negative power.
 
 
 
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