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Integration by Parts

  1. Oct 14, 2015 #1
    The Integral:
    [tex]\int{\sin{(\theta)}*\cos{(\theta)}*d\theta}[/tex]
    Attempt to solve by Integration by Parts:
    [tex]\int{u*dv} = u*v - \int{v*du}[/tex]
    [tex]u = \sin{(\theta)}[/tex]
    [tex]du = \cos{(\theta)}*d\theta[/tex]
    [tex]v = \sin{(\theta)}[/tex]
    [tex]dv = \cos{(\theta)}*d\theta[/tex]
    Bringing back to the beginning.
     
  2. jcsd
  3. Oct 14, 2015 #2

    jedishrfu

    Staff: Mentor

    Have you considered using the ##sin(2\theta) = 2 sin(\theta) cos(\theta)## identity or do you need to use parts?

    Also what if you add ##udv + vdu## integrals together to be = ## u . v ##?
     
  4. Oct 14, 2015 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    In fact, I wouldn't use the that identity! Since [itex](sin(\theta)'= cos(\theta)[/itex], the substitution [itex]u= sin(\theta)[/itex] so [itex]du= cos(\theta)d\theta[/itex] gives [itex]\int sin(\theta)cos(\theta) d\theta= \int u du= \frac{1}{2}u^2+ C[/itex].

    But you can do this "by parts" using [itex]u= sin(\theta)[/itex], [itex]dv= cos(\theta)d\theta[/itex] as you suggest. [itex]du= cos(\theta)d\theta[/itex] and [itex]v= sin(\theta)d\theta[/itex] so the integral becomes [itex]\int sin(\theta)cos(\theta)d\theta= sin^2(\theta)- \int sin(\theta) cos(\theta) d\theta[/itex].

    Now, that is NOT "back to the beginning" because we can add [itex]\int sin(\theta)cos(\theta) d\theta[/itex] to both sides to gbet [itex]2 \int sin(\theta)cos(\theta)d\theta= sin^2(\theta)[/itex] and dividing both sides by 2 gives the previous result.
     
    Last edited by a moderator: Oct 14, 2015
  5. Oct 14, 2015 #4
    HallsofIvy has made a good point. Always write down the LHS and the RHS of the equation while integrating by parts when you're dealing with functions which are related to their first few derivatives with a constant (chiefly exponential and trigonometric functions).
     
  6. Oct 14, 2015 #5

    Mark44

    Staff: Mentor

    What HallsOfIvy and PWiz are talking about is using integration by parts and algebraically solving for an integral. Here's an example.
    ##\int e^x \cos x dx## -- int. by parts, with ##u = e^x, dv = \cos x dx##, so ##du = e^x, v = \sin x##
    ##= e^x \sin x - \int e^x \sin x dx## -- int. by parts again, with ##u = e^x, dv = -\sin x dx## and ##du = e^x dx, v = \cos x##
    ##= e^x \sin x + e^x \cos x - \int e^x \cos x dx##

    It looks like we've come full circle here, but actually we haven't. The above shows that
    ##\int e^x \cos x dx = e^x \sin x + e^x \cos x - \int e^x \cos x dx##
    Add ##\int e^x \cos x dx## to both sides of the equation above, and solve algebraically for this integral. Of course, you need to add the arbitrary constant in your final answer.
     
  7. Oct 14, 2015 #6

    phion

    User Avatar
    Gold Member

    You don't need to use integration by parts. Try a simple u-substitution.
     
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