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Core Maths 4: Integration/Trig/double and triple angle formulae

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    Hi there

    Please could someone take a look at this and advise if I am on the right lines, and maybe give me a little nudge in the right direction to solve the question



    Many thanks
    Jackie
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    No you're going down the wrong route.

    Express cos^3(x) as cos(x)cos^2(x) and go from there. If you're still stuck, I'll help you some more
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    (Original post by jackie11)
    Hi there

    Please could someone take a look at this and advise if I am on the right lines, and maybe give me a little nudge in the right direction to solve the question



    Many thanks
    Jackie
    You are correct till  \cos^3 A - 3 \sin^2 A \cos A

    Then how did you wrote,  \cos^3 A = 3 \sin^2 A \cos A \ ?

    What you have is  \cos (3A) =  \cos^3 A - 3 \sin^2 A \cos A  \implies \cos^3 A = \cos (3A) + 3 \sin^2 A \cos A


     \displaystyle \int \cos^3 x \ dx = \int \left( \cos (3x) + 3 \sin^2 x \cos x \right) \ dx

    Hint for integrating  3 \sin^2 x \cos x
    Spoiler:
    Show

    It is an integral of the form  \displaystyle \int f'(x) f(x) \ dx

    So try differentiating  \sin^3 x
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    Another approach can be to express  \cos^3 x as  \cos x \cos^2 x = \cos x ( 1- \sin^2 x ) = \cos x - \cos x \sin^2 x
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    (Original post by Lord of the Flies)
    It seems you have taken the complicated route.

    Try considering \displaystyle\int_{-\pi/2}^{\pi/2}\cos x\cos^2 x by parts
    I think your route is also complicated.

    The method that Lukas gave is the best way to tackle this question.
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    (Original post by Lord of the Flies)
    What method?

    Also, I don't see how an integration by parts is more complicated than the stream of equations in the OP:

    Without difficulty, by parts twice gives:

    \displaystyle\int\cos x\cos^2 x=[\sin x \cos^2]-2\int\sin^2 x\cos x+[\sin^3x]+2\int\sin^2x\cos x

    =[\sin x \cos^2]+[\sin^3x]
    I don't mean the OP's method. I mean the method given by lukas1051:

    \displaystyle \cos^3 x = \cos x (1-\sin^2 x) = \cos x -\cos x \sin^2 x

    Both of these integrals are easy. It is far quicker and "nicer" to do this rather than use integration by parts twice.
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    (Original post by notnek)
    \displaystyle \cos^3 x = \cos x (1-\sin^2 x) = \cos x -\cos x \sin^2 x

    It is far quicker and "nicer" to do this rather than use integration by parts.
    Yes, just realised this... :oops:

    Besides I made an idiotic mistake in my post so I'm going to leave now...
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    (Original post by raheem94)
    You are correct till  \cos^3 A - 3 \sin^2 A \cos A

    Then how did you wrote,  \cos^3 A = 3 \sin^2 A \cos A \ ?

    What you have is  \cos (3A) =  \cos^3 A - 3 \sin^2 A \cos A  \implies \cos^3 A = \cos (3A) + 3 \sin^2 A \cos A

    [/spoiler]
    yes you are right, omg what a silly mistake to make haha
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    ok I am still not getting the right answer for this, am I on the right lines?

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    (Original post by jackie11)
    ok I am still not getting the right answer for this, am I on the right lines?
    \displaystyle\int \cos x\sin^2 x\;dx should remind you of something...

    Hint:

    Spoiler:
    Show

    =\dfrac{1}{3}\displaystyle\int 3\cos x\sin^2 x dx
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    (Original post by jackie11)
    ok I am still not getting the right answer for this, am I on the right lines?
    You main mistake is in saying that

     \displaystyle \int \cos x \sin^2 x \ dx = \int cos x \ dx \times \int \sin^2 x \ dx

    You can't split up a product in an integral like this. In general,

    \int AB \neq \int A \times \int B

    Instead, you need to integrate \cos x \sin^2 x as one expression.

    Try differentiating \sin^3 x and see what you get. This should give you a big clue as to what the integral of \cos x \sin^2 x is.
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    Using the chain rule, differentiating sin³ x gives me;

    -3 sin² x cos x

    So the integral of cos x sin² x is -1/3 sin³ x.

    Am I on the right lines now? lol
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    (Original post by jackie11)
    Using the chain rule, differentiating sin³ x gives me;

    -3 sin² x cos x

    So the integral of cos x sin² x is -1/3 sin³ x.

    Am I on the right lines now? lol
    Almost - you made a sign mistake. The derivative of sin is cos, not -cos.
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    (Original post by jackie11)
    Using the chain rule, differentiating sin³ x gives me;

    -3 sin² x cos x

    So the integral of cos x sin² x is -1/3 sin³ x.

    Am I on the right lines now? lol
    Almost, but the derivative of \sin^3 x shouldn't have a negative sign since the derivative of \sin x is \cos x not -\cos x.
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    ok, yes I have it!!! woohoo !!!!!!!!!!

    Thank you both so much.

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