Evaluate the Integral

Discussion in 'Calculus' started by iwntchckknnggts, Nov 28, 2021.

  1. iwntchckknnggts

    iwntchckknnggts

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    i need help
     

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    iwntchckknnggts, Nov 28, 2021
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  2. iwntchckknnggts

    MathLover1

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    int(sin(x)*ln(1+sin(x))dx) (I am using int() instead symbol ∫)

    apply integration by parts:

    Using the product rule, we have that

    int(udv)=uv-int(vdu)

    let u=ln(1+sin(x)) and v=sin(x)dx

    Then du=(ln(sin(x)+1))′dx=(cos(x)/(sin(x)+1))dx
    and

    v=int(sin(x)dx)=-cos(x)

    The integral can be rewritten as

    int(ln(sin(x)+1)sin(x)dx)
    =(ln(sin(x)+1)*(-cos(x))-int(-cos(x))*(cos(x)/(sin(x)+1))dx)
    =(-ln(sin(x)+1)cos(x)-int((-cos^2(x)/(sin(x)+1))dx)

    Rewrite the cosine in terms of the sine, rewrite the numerator further, use the formula for difference of squares, and simplify:

    -ln(sin(x)+1)cos(x)- int((-cos^2(x)/(sin(x)+1))dx)
    =-ln(sin(x)+1)cos(x)- int((sin(x)-1)dx)

    Integrate term by term:

    -ln(sin(x)+1)cos(x)- int((sin(x)-1)dx)
    =-ln(sin(x)+1)cos(x)-(-int(1dx)+int(sin(x)dx)

    Apply the constant rule int(c)dx=cx with c=1:

    -ln(sin(x)+1)cos(x)-int(sin(x)dx)+int(1dx)
    =-ln(sin(x)+1)cos(x)-int(sin(x)dx)+x

    The integral of the sine is int(sin(x)dx)=-cos(x):

    x-ln(sin(x)+1)cos(x)-int(sin(x)dx)=x-ln(sin(x)+1)cos(x)-(-cos(x))

    Therefore,

    int(ln(sin(x)+1)sin(x)dx)=x-ln(sin(x)+1)cos(x)+cos(x)

    Add the constant of integration:

    int(ln(sin(x)+1)sin(x)dx)=x-ln(sin(x)+1)cos(x)+cos(x)+C
    Answer:
    int(ln(sin(x)+1)sin(x)dx)=x-ln(sin(x)+1)cos(x)+cos(x)+C
     
    MathLover1, Nov 28, 2021
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    nycmathguy and iwntchckknnggts like this.
  3. iwntchckknnggts

    iwntchckknnggts

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    THANK YOU SO MUCH
     
    iwntchckknnggts, Nov 28, 2021
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  4. iwntchckknnggts

    nycmathguy

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    Can you please use the math symbols method for easy reading? You know, the wolfram stuff.
     
    nycmathguy, Nov 28, 2021
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  5. iwntchckknnggts

    MathLover1

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    I was trying, for some reason it didn't work
     
    MathLover1, Nov 28, 2021
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    nycmathguy likes this.
  6. iwntchckknnggts

    nycmathguy

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    Ok. Thanks anyway.
     
    nycmathguy, Nov 28, 2021
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