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Math Help - Definite Integrals!

  1. #1
    Senior Member polymerase's Avatar
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    Definite Integrals!

    \int^1_0 arcsin\left(\frac{2x}{1+x^2}\right)

    PS. How do you get the brackets to enclose the entire expression?
    Last edited by polymerase; January 21st 2008 at 07:34 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by polymerase View Post
    PS. How do you get the brackets to enclose the entire expression?
    well, i can answer that part

    type "\left(" and "\right)"
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  3. #3
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    Quote Originally Posted by polymerase View Post
    \int^1_0 arcsin\left(\frac{2x}{1+x^2}\right)
    I was tryin' to find a nice way to tackle this but I couldn't.

    Looks like it's pretty straightforward.

    Integrate by parts:

    u = \arcsin \frac{{2x}}<br />
{{1 + x^2 }} \implies du =   \frac{2}<br />
{{1 + x^2 }}\,dx & dv=dx, so

    \int_0^1 {\arcsin \frac{{2x}}<br />
{{1 + x^2 }}\,dx}  = \left. {x\arcsin \frac{{2x}}<br />
{{1 + x^2 }}} \right|_0^1  - \int_0^1 {\frac{{2x}}<br />
{{1 + x^2 }}\,dx} .

    The rest follows.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Integrate by parts:
    that's what i thought...but i was too lazy to check...
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  5. #5
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Krizalid View Post
    I was tryin' to find a nice way to tackle this but I couldn't.
    I know you dont like this, but you could use the substitution u=tan\;x then use by parts on \int 2u\;sec^2\;u\;du
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  6. #6
    Math Engineering Student
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    Anyway you can't get rid of integration by parts.

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