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Math Help - Easy definite integral problem

  1. #1
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    Easy definite integral problem

    Easy definite integral problem-codecogseqn.gif

    So, i use substitution sinx=t, cosxdx=dt. When i put it in, i got ln(t)dt, now i know i have to use partial integration, but how do I set limits?
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  2. #2
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    Re: Easy definite integral problem

    Quote Originally Posted by DonnieDarko View Post
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    So, i use substitution sinx=t, cosxdx=dt. When i put it in, i got ln(t)dt, now i know i have to use partial integration, but how do I set limits?
    \int {\ln (t)dt}  = t\ln (t) - t
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  3. #3
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    Re: Easy definite integral problem

    Hello, DonnieDarko!

    \int^{\frac{2\pi}{3}}_{\frac{\pi}{6}} \cos x\ln(\sin x)\,dx

    \text{So, I use substitution: }\:t \,=\, \sin x,\;dt\,=\,\cos x\,dx

    \text{When I put it in, i got: }\:\int \ln t\,dt

    \text{Now I know I have to use partial integration, but how do I set limits?}

    Plug the x-limits into your substitution.

    You have: . t \:=\:\sin x

    \text{When }x = \tfrac{\pi}{6}\!:\;\;t \:=\:\sin\tfrac{\pi}{6} \:=\:\tfrac{1}{2}

    \text{When }x = \tfrac{2\pi}{3}\!:\;\;t \:=\:\sin\tfrac{2\pi}{3} \:=\:\tfrac{\sqrt{3}}{2}


    \displaystyle\text{The integral becomes: }\int^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}}\!\! \ln t\,dt
    Thanks from DonnieDarko
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    Re: Easy definite integral problem

    Thank you very much Soroban, wasnt sure about it, so its always good to check
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    Re: Easy definite integral problem

    I assume that by "partial integration", you mean what I would call "integration by parts":
    \int_a^b udv= \left[uv\right]_a^b- \int_a^b vdu

    Here, once you have made the substitution, t= sin(x), and have [itex]\int ln(t)dt[/itex], let u= ln(t), dv= dt. Then du= (1/t)dt and v= t so we have
    \left[t ln(t)\right]_{1/2}^{\sqrt{3}/2}- \int_{1/2}^{\sqrt{3}/2} dt= \left[tln(t)- t\right]_{1/2}^{\sqrt{3}/2}
    That's how Plato got his answer.
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