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Core Maths 3: How do you integrate this?

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    Hi there

    I have the question:

    \displaystyle \int sin \frac{x}{3} dx

    I know that sin x integrated is - cos x,

    so is the answer to my question, before working out the exact values with the limits, this;

    \displaystyle - cos \frac{x}{3} + c

    Many thanks
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    The integral of sin(ax+b) is -(1/a)cos(ax+b).
    That means in this situation a is 1/3.
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    ok so in my case its -3 cos x ??
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    The x value must remain the same as 'a' remains in front it.
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    so its -3 cos 1/3x which is - cos x??
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    Yes and no. Cos(ax) is one value and 'a' cannot be changed by the value cos is multiplied by. The answer is -3cos(x/3). There's also a '+ c' but you already knew that. :P
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    ok thank you very much
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    (Original post by jackie11)
    ok thank you very much
    No problem; thanks for the rep.
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    Hey
    I would suggest using the substitution method.
     \frac{x}{3} = u therefore \frac{du}{dx} = \frac{1}{3} so dx = 3 du

    The original equation is  \displaystyle \int sin \frac{x}{3}dx if you put in all the substituions you should get
     \displaystyle \int sin u 3du
    which simplifies to  \displaystyle 3 \int sin u du

    This should be fine to solve and then replace u with x/3 to get the final answer.

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