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    Could someone help me with this question please:

    How would i integrate  sin^2xcos^2x?
    This is Question 1h Exercise 6C from C4 book.
    ................................ ................................ .....
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    Well you do need to use double angle formula but you can just reapply to deal with \cos^2 2x

    \cos 4x = 2 \cos^2 2x -1
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    (Original post by nm786)
    Could someone help me with this question please:

    How would i integrate  sin^2xcos^2x?
    This is Question 1h Exercise 6C from C4 book.

    This is what i did:
    cos2x=1-2sin^2x and cos2x=2cos^2x-1
    So, sin^2x=-1/2(cos2x-1) and cos^2x=1/2(cos2x+1)
    multiplying both of these: -1/4(cos^22x-1)
    what should i do next? I'm confused.
    \displaystyle \sin^2 x \cos^2 x = \frac{\sin^2 2x}{4}
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    (Original post by nerak99)
    Well you do need to use double angle formula but you can just reapply to deal with \cos^2 2x

    \cos 4x = 2 \cos^2 2x -1
    Thanks for replying mate, so sould i sub this into the last equation and and integrate?: 1/2(1/2(cos4x+1)+1)
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    Well, working from what you did you could do that (I have not checked your working) but what Mr M showed you is a more elegant way of working if you want to take it from the top.
    You need to use \cos 4x = 1 - 2 \sin^2 2x

    Mr M has use \sin 2x = 2 \sin x \cos x and squared it.
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    (Original post by nerak99)
    Well, working from what you did you could do that (I have not checked your working) but what Mr M showed you is a more elegant way of working if you want to take it from the top.
    You need to use \cos 4x = 1 - 2 \sin^2 2x

    Mr M has use \sin 2x = 2 \sin x \cos x and squared it.
    so from what i did: 1/2(1/2(cos4x+1)+1)
    =1/2(1/2cos4x+1/2+1)
    =1/4cos4x+1/4+1/2
    integrating above = 1/4*1/4sin4x+1/4x+1/2x+c
    =1/16sin4x+3/4x+c
    but the answer is 1/8x-1/32sin4x+c?

    I'll use the equation Mr M suggested in a minute.
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    (Original post by Mr M)
    \displaystyle \sin^2 x \cos^2 x = \frac{\sin^2 2x}{4}
    I understand how you got that now but how would i go about integrating it?
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    (Original post by nm786)
    I understand how you got that now but how would i go about integrating it?
    You should know that \displaystyle \sin^2 \theta = \frac{1-\cos 2\theta}{2}
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    (Original post by Mr M)
    You should know that \displaystyle \sin^2 \theta = \frac{1-\cos 2\theta}{2}
    i never knew that
    what about cos^2x?
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    (Original post by nm786)
    i never knew that
    what about cos^2x?
    It is a key fact in Core 4 integration.

    \displaystyle \cos^2 \theta= \frac{1 + \cos 2\theta}{2} but that is not relevant to this question.
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    (Original post by Mr M)
    It is a key fact in Core 4 integration.

    \displaystyle \cos^2 \theta= \frac{1 + \cos 2\theta}{2} but that is not relevant to this question.
    alright i get it this is from cos2x identity right?
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    (Original post by nm786)
    alright i get it this is from cos2x identity right?
    how would i do this question sir,
    using what you said: sin^22x/4=sin^2xcos^2x
    =(2sinxcosx)^2=1/2sin^2xcos^2x
    using the identitiessin^2x=-1/2(cos2x-1) and cos^2x=1/2(cos2x+1) multiplying them together and 1/2:
    =(1/2)[(-1/4(cos^22x-1)]
    =-1/8cos^22x+1/8
    I can use -1/2(cos2x-1) again so -1/8(-1/2cos2x-1)+1/8
    =1/16cos2x+1/8+1/8=116cos2x+2/8
    integrating above i get (1/2)(1/16)sin2x+2/8x i have no idea where i went wrong
    You make things so complicated!

    \displaystyle \int \sin^2 x \cos^2 x \, dx = \int \frac{\sin^2 2x}{4} \, dx = \int \frac{1-\cos 4x}{8} \, dx
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    (Original post by Mr M)
    You make things so complicated!

    \displaystyle \int \sin^2 x \cos^2 x \, dx = \int \frac{\sin^2 2x}{4} \, dx = \int \frac{1-\cos 4x}{8} \, dx
    i don't understand how you went from \int \frac{\sin^2 2x}{4} \, dx = \int \frac{1-\cos 4x}{8} \, dx
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    (Original post by nm786)
    i don't understand how you went from \int \frac{\sin^2 2x}{4} \, dx = \int \frac{1-\cos 4x}{8} \, dx
    Post 9?
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    (Original post by Mr M)
    Post 9?
    oh yeah sorry,
    so  \int \frac{1-\cos 4x}{8} \, dx = \int 1/8-\frac{cos4x}{8} \, dx
    =1/8-(1/8)(1/4)sin4x+c
    EDIT:=1/8x-1/32sin4x+C which is the right answer!
    Cheers Sir!
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    (Original post by nm786)
    oh yeah sorry,
    so  \int \frac{1-\cos 4x}{8} \, dx = \int 1/8-\frac{cos4x}{8} \, dx
    =1/8-(1/8)(1/4)sin4x+c
    =1/8-1/32sin4x+C which is the right answer!
    Cheers Sir!
    \displaystyle \int \frac{1}{8} \, dx = \frac{x}{8}+k
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    (Original post by Mr M)
    \displaystyle \int \frac{1}{8} \, dx = \frac{x}{8}+k
    damn it i forgot the x
    Much appreciated for the help sir!
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    (Original post by nm786)
    damn it i forgot the x
    Much appreciated for the help sir!
    You are welcome.
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    (Original post by nm786)
    alright i get it this is from cos2x identity right?
    how would i do this question then sir,
    using what you said: sin^22x/4=sin^2xcos^2x
    =(2sinxcosx)^2=1/2sin^2xcos^2x
    do you think to
    \frac{(2\sin x\cos x)^2}{4}=\frac{4\cdot \sin^2 x \cos^2 x}{4}=\sin^2 x\cdot \cos^2 x
    here
    using the identitiessin^2x=-1/2(cos2x-1) and cos^2x=1/2(cos2x+1) multiplying them together and 1/2:
    There would not be 'and 1/2'
    =(1/2)[(-1/4(cos^22x-1)]
    =-1/8cos^22x+1/8
    That is without 1/2 factor
    \frac{1-\cos^2 2x}{4}=\frac{\sin^2 2x}{4}
    which is same that was said to you.
    I can use -1/2(cos2x-1) again so -1/8(-1/2cos2x-1)+1/8
    Just little difference only
    Consider that  \cos (2\cdot 2x)=\cos 4x
    (and +1/4 at the end)
    \frac{1-\cos^2 2x}{4}=\frac{1}{4}-\frac{1}{4}\cos^2 2x=
    =\frac{1}{4}-\frac{1}{8}(1+\cos 4x)=\frac{1}{8}-\frac{1}{8}\cos 4x
    =1/16cos2x+1/8+1/8=116cos2x+2/8
    integrating above i get (1/2)(1/16)sin2x+2/8x +C i have no idea where i went wrong the answer is something else
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    (Original post by ztibor)
    do you think to
    \frac{(2\sin x\cos x)^2}{4}=\frac{4\cdot \sin^2 x \cos^2 x}{4}=\sin^2 x\cdot \cos^2 x
    here

    There would not be 'and 1/2'

    That is without 1/2 factor
    \frac{1-\cos^2 2x}{4}=\frac{\sin^2 2x}{4}
    which is same that was said to you.

    Just little difference only
    Consider that  \cos (2\cdot 2x)=\cos 4x
    (and +1/4 at the end)
    \frac{1-\cos^2 2x}{4}=\frac{1}{4}-\frac{1}{4}\cos^2 2x=
    =\frac{1}{4}-\frac{1}{8}(1+\cos 4x)=\frac{1}{8}-\frac{1}{8}\cos 4x
    He finished the question 25 minutes ago.

 
 
 
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