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I am stumped! Solving an indefinite Integral

  1. Apr 30, 2007 #1
    Problem: [int]cosx(sinx)dx
    Given: x=pi; f(pi)=13.4

    I am utterly confused on how to solve this integral. I am 99% positive (which is nothing in the math world) that I need to apply the product rule to all of this in order to find the antiderivative. However, no matter how I think of going about it, I will always have that addition sign in there that ruins the whole equation so I can't divide/null particular variables in order to end up with the cosx(sinx) as the derivative.

    What I have done so far is firstly finding the antiderivatives for cosx and sinx. I know cosx=sinx + C and sinx=-cosx + C. But now I do not know the next step. I have solved other integral problems like e^2x and the like, so it's not a foreign concept, but with this one I have no clue.

    I have thought about using the sum and difference identities for cos and sin:

    sin2x=sinx(cosx) + sinx(cosx)
    cos2x=(cosx)^2 - (sinx)^2

    Although once again, I couldn't figure out how those could help me, in addition to applying the quotient rule... however, like I said previously, I am almost positive I need to be using the product rule to solve this integral function, but then what do I do? I know some of you have suggestions, please help, this is driving me mad!
    Last edited: Apr 30, 2007
  2. jcsd
  3. Apr 30, 2007 #2
    cos(x)sin(x)=1/2 sin(2x)
  4. Apr 30, 2007 #3
    Is this the antiderivative or is that actually what cosx(sinx) is equivalent to?

    Epiphany: OH!!! Brilliant, I see it now. ha.
    Last edited: Apr 30, 2007
  5. Apr 30, 2007 #4

  6. Apr 30, 2007 #5
    Yeh, thanks, just took me several seconds to figure out the implied steps it took to get there...

    Is this right?:

    1/2[int]sin(2x)dx = 1/2(-cos(2x)) + C => 13.4 = 1/2(-cos(2pi)) + C => 13.4 - 1/2(-1) = C => 13.9 = C
  7. Apr 30, 2007 #6


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    Another way to do this- the way I would have done it- would be to let u= sin(x). Then du= cos(x)dx.
  8. Apr 30, 2007 #7
    I hadn't thought of chunking...

    Well, I can try that way if the way I did it was false... so did I answer the integral correctly or must I try again?
  9. May 1, 2007 #8

    Gib Z

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    They way you did it is incorrect, and it seems you may need a brushup on your integration. The integral of sin x may be negative cos x, but the integral of sin (2x) is not negative cos (2x).

    Remember to check the answer the integrals derivative must be the original problem, which it isn't. You must try again.

    To do it christianjb's way, [tex]\frac{1}{2}\int \sin (2x) dx[/tex] you would let u=2x, from that we get du=2 dx, dx= du/2. Substitute them in:

    [tex]\frac{1}{4}\int \sin (u) du[/tex].

    Since the integral of sin x is -cos x, that integral evaluates to [tex]\frac{-1}{4} \cos (2x) + C[/tex] because you substitute u=2x back in.

    The way I would have done it, and that Halls pointed out is that since we already know (sin x) ' = cos x,

    [tex]\int \sin x \cos x dx [/tex]

    u=sin x
    du=cos x dx
    [tex]\int u du = \frac{u^2}{2} + C = \frac{\sin^2 x}{2}+C[/tex].

    You could also have done u= cos x, however its quicker because the signs of the derivatives match up in the case of u=sin x.

    If you do use u= cos x, you will get what appears to be a different answer.

    [tex]\frac{\cos^2 x}{2} + C[/tex]

    Do not be alarmed, they are both equal. since [itex]\sin^2 x + \cos^2 x = 1[/itex], the 2 different anti derivatives differ by 1/2, which is a constant.
  10. May 1, 2007 #9
    Thanks alot, that makes sense in that way, but I ended up figuring out that I needed to apply the chain rule to the remainder of my initial thought process. Again, thanks alot for the mental push :-)
  11. May 1, 2007 #10


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    Whoops, you make a minor mistake here, you forgot the minus sign. o:). The correct answer should be:
    [tex]\int \sin x \cos x dx = - \int u du = - \frac{u ^ 2}{2} + C = - \frac{\cos ^ 2 x}{2} + C[/tex], where we use the u-substitution: u = cos x.
  12. May 1, 2007 #11
    My final process included the final equation, f(pi) = 1/4-cos(2pi) + C = 13.4; consequently, C = 13.65. In order to check my work of the antiderivative, I applied the chain rule to this original function. Consequently, f(x)' = 1/4(sin2x)(2) => 1/2sin(2x) or sinxcosx, which refers back to the function included in the integral. I really hope that is right or am I just fooling myself...

    I am actually doing this independently, in addition to my Calculus class, so it is OK if you feel you need to nudge me along toward the correct answer. This is not exactly homework participation, just extracurricular.
  13. May 2, 2007 #12

    Gib Z

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    Your antiderivative is correct.

    However I am a bit confused, is f(x) what you want the area under, or f'(x)?
  14. May 2, 2007 #13
    I just labelled f(x) as the original function (the antiderivative) and f(x)' is the derivative function in the integral, sinxcosx. Was that your misunderstanding?

    And actually, the reason I did not understand what the d and du variables were, is because we just went over their uses today in class. So thanks for giving me the heads up, Gib Z.
  15. May 3, 2007 #14

    Gib Z

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    Yea that was my misunderstanding, thats. Usually people will call the function they want to integrate f(x) and the anti derivative F(x). That is because sometimes when doing integrals f'(x) must be known, and it becomes lame to write f''(x).
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