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Math Help - Trig Integration

  1. #1
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    Trig Integration

    Hello, the book answer for this is ln2 or .693

    They let u be 1+sint and go from there.
    I did it another way but am gettin the wrong answer (which is 1)

    ThanksTrig Integration-imageuploadedbytapatalk-21370042294.793468.jpg
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  2. #2
    Junior Member ReneG's Avatar
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    Re: Trig Integration

    The first step was your mistake. \frac{\cos t}{1+\sin t} \neq \cos t + \frac{\cos t}{\sin t}

    So you're given \int_0^\frac{\pi}{2} \frac{\cos t}{1 + \sin t} \, \mathrm{d}x

    It would be better to set u equal to the denominator since the derivative of 1 is 0, allowing you to substitute more.

    Set u as the denominator u = 1 + \sin t
    \mathrm{d}u = \cos t \, \mathrm{d}t
    Substitute \int_0^\frac{\pi}{2} \frac{1}{u} \, \mathrm{d}u
    Integrate \ln u \, \biggr|_0^\frac{\pi}{2}
    Back track u \ln |1 + \sin t \, | \, \biggr|_0^\frac{\pi}{2}
    Evaluate anti-derivative at limits \ln |1 + \sin \frac{\pi}{2} \, | - \ln |1 + \sin 0 |
    Simplify 2
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  3. #3
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    Re: Trig Integration

    Quote Originally Posted by ReneG View Post
    The first step was your mistake. \frac{\cos t}{1+\sin t} \neq \cos t + \frac{\cos t}{\sin t}

    So you're given \int_0^\frac{\pi}{2} \frac{\cos t}{1 + \sin t} \, \mathrm{d}x

    It would be better to set u equal to the denominator since the derivative of 1 is 0, allowing you to substitute more.

    Set u as the denominator u = 1 + \sin t
    \mathrm{d}u = \cos t \, \mathrm{d}t
    Substitute \int_0^\frac{\pi}{2} \frac{1}{u} \, \mathrm{d}u
    Integrate \ln u \, \biggr|_0^\frac{\pi}{2}
    Back track u \ln |1 + \sin t \, | \, \biggr|_0^\frac{\pi}{2}
    Evaluate anti-derivative at limits \ln |1 + \sin \frac{\pi}{2} \, | - \ln |1 + \sin 0 |
    Simplify 2

    When you make substitution you must change the limits For example
    u = 1+sin t
    When t = 0 u = 1 and when t = pi/2 , u = 2
    Thus the integral would be from 1 to 2, intgreal 1/u du
    This will avoid some steps and would help in getting solutin early. in this case it would be simply log 2 since log 1 = 0
    Thanks from ReneG
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  4. #4
    Junior Member ReneG's Avatar
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    Re: Trig Integration

    Quote Originally Posted by ibdutt View Post
    When you make substitution you must change the limits For example
    u = 1+sin t
    When t = 0 u = 1 and when t = pi/2 , u = 2
    Thus the integral would be from 1 to 2, intgreal 1/u du
    This will avoid some steps and would help in getting solutin early. in this case it would be simply log 2 since log 1 = 0
    Thanks for the heads up. I nice intuitive trick.
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