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Which method of integration for sin^2(x) dx

  1. Jul 4, 2012 #1
    During maths class last semester this integral came up in the course of discussion and my lecturer gave a quick outline of how to solve it but I didn't grasp it at the time and we moved on. I'd like to know how to do it though! The integral is:

    [itex]\int sin^{2}(x)[/itex]

    The next step was:

    [itex]\int sin(x).(-cos(x)')[/itex]

    where [itex]-cos(x)'[/itex] denotes the derivative of [itex]-cos(x)[/itex] and the implication was that [itex]sin^{2}(x)[/itex] is the result of applying the chain rule to whatever compound function it is the derivative of. I can't remember the result but further steps were skipped and he jumped straight to the answer at this point.

    I'm not sure how to proceed here, because the chain rule implies that [itex]\frac{dy}{dx} f(g(x)) = f'(g(x) . g'(x)[/itex], and in my example [itex]g(x) = x[/itex] and [itex]g'(x) = 1[/itex] which doesn't help me end up with [itex]sin(x)[/itex] as required.

    Could anyone poke me in the right direction to solve this? =)
  2. jcsd
  3. Jul 4, 2012 #2


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    I'd forget the chain rule and try to re-write [itex]\sin^2x[/itex] as something I could integrate, if I were you. It might be possible by parts, too, but I haven't actually tried that so I wouldn't swear to it.
  4. Jul 4, 2012 #3


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    The method I was taught was to use the double angle formula for cosine.
    I really have no idea what your teacher is trying to do.
  5. Jul 4, 2012 #4

    D H

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    You don't want the chain rule. You want the product rule, the integration equivalent of which is integration by parts.

    Before nudging you in the right direction, there is a much easier way to find [itex]\int sin^2x\,dx[/itex]. Use the identity [itex]\sin^2x = (\frac{1-\cos(2x)} 2[/itex]. The right hand side is easy to integrate.

    The product rule is [itex](f(x)g(x))' = f'(x)g(x) + f(x)g'(x)[/itex]. Transforming to integration,[tex]\int (f(x)g(x))' dx = f(x)g(x) = \int f'(x)g(x)dx + \int f(x)g'(x)dx[/tex]
    Note that this is integration by parts, just written a bit differently. You are using [itex]f(x)=\sin x[/itex] and [itex]g(x)=-\cos x[/itex]. Plug these in. You'll eventually need a different trig identity, [itex]\sin^2x + \cos^2x = 1[/itex], to finish it off.
    Last edited: Jul 5, 2012
  6. Jul 5, 2012 #5
    Thanks for the replies. I'm really awful at remembering trig identities (and remembering in general =S), I had a large gap between high school and university maths, so I have to re-learn a lot of the basics as I go along.

    I solved it by using the double angle identity [itex]sin^{2}x = \frac{1 - cos(2x)}{2}[/itex] and integration by substitution - [itex]u = 2x, du = 2dx, dx = \frac{1}{2}du[/itex], and my result was [itex]\frac{x}{2} - \frac{1}{4}sin(2x)[/itex]. I took the derivative to check my answer and it worked, so that's great!

    I'm going to have a go at the integration by parts method now.
  7. Jul 5, 2012 #6

    D H

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    Adyssa, I had a typo (matho) in my previous post.

    It is always a good idea to double check your work. Good job!
  8. Jul 7, 2012 #7
    You can try another technique...
    I=∫sin[itex]^{2}[/itex]xdx ---------(i)
    or I=∫dx-∫cos[itex]^{2}[/itex]xdx
    or I=x-∫cos[itex]^{2}[/itex]xdx +c[itex]_{1}[/itex] -------(ii)

    Adding (i) and (ii),
    2I=x-∫[cos[itex]^{2}[/itex]x-sin[itex]^{2}[/itex]x]dx +c[itex]_{1}[/itex]
    2I=x-∫cos2xdx +c[itex]_{1}[/itex]
    2I=x-1/2 sin2x +c
    I=x/2 - 1/4 sin2x+c
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