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Intergration by parts for sin(x)cos(x)

  1. May 30, 2014 #1
    I know its easier to use the substitution method, by I'm trying to see how it'll work for integration by parts. I follow the LIATE method for integration by parts.


    Now if I take u=cos(x) and dv = sin(x), the answer changes.

    Can you please explain this to me? Which is the 'right' answer?

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  2. jcsd
  3. May 30, 2014 #2
    Rather than answer your question directly, let me provide another example of the same phenomenon that might make it easier to identify why both answers are correct.

    If you compute ##\int (x+1)\ dx## using the sum rule for antiderivatives and the "reverse power rule", you get $$\int x+1\ dx=\frac{1}{2}x^2+x+C.$$ If instead you use a ##u##-sub, with ##u=x+1##, you get $$\int (x+1)\ dx=\int u\ du=\frac{1}{2}u^2+C=\frac{1}{2}(x+1)^2+C.$$

    Now the exact same thing is happening in your example as is happening here; you have two different-looking answers that are both correct. How can that be?
  4. May 30, 2014 #3

    D H

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    And closer to home, let's do ##\int \sin(x) \cos(x) dx## a third way, using the trig substitution ##\sin(2x)=2\sin(x) \cos(x)##. Thus ##\int \sin(x) \cos(x) dx = \int \frac {\sin(2x)}2 dx = -\frac {\cos(2x)} 4 + C##.

    To quote gopher_p,
  5. May 30, 2014 #4
    dear friend , all these answers are different only by a constant, so they are the same answers.
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