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Math Help - Calculating intregral

  1. #1
    fkf
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    Calculating intregral

    I've been trying to calculating the following integral
    \int xsin^{-1}(x)\, dx

    Here's my solution as far as I've been able to calculate. Let
    U=sin^{-1}x \Rightarrow dU=\frac{1}{\sqrt{1-x^2}}dx
    dV = x \cdot dx \Rightarrow V = \frac{x^2}{2}
    Then
    I=VU-\int V\cdot du=sin^{-1}x \cdot \frac{x^2}{2} -\int \frac{x^2}{2}\cdot \frac{1}{\sqrt{1-x^2}}dx

    I get stuck here, using U and V for new functions and using the relationship again doesn't lead me anywhere.
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  2. #2
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    Re: Calculating intregral

    Quote Originally Posted by fkf View Post
    I've been trying to calculating the following integral
    \int xsin^{-1}(x)\, dx

    Here's my solution as far as I've been able to calculate. Let
    U=sin^{-1}x \Rightarrow dU=\frac{1}{\sqrt{1-x^2}}dx
    dV = x \cdot dx \Rightarrow V = \frac{x^2}{2}
    Then
    I=VU-\int V\cdot du=sin^{-1}x \cdot \frac{x^2}{2} -\int \frac{x^2}{2}\cdot \frac{1}{\sqrt{1-x^2}}dx

    I get stuck here, using U and V for new functions and using the relationship again doesn't lead me anywhere.
    You can see how it works here.

    Ckick the show steps tab.
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  3. #3
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    Re: Calculating intregral

    Quote Originally Posted by fkf View Post
    I've been trying to calculating the following integral
    \int xsin^{-1}(x)\, dx

    Here's my solution as far as I've been able to calculate. Let
    U=sin^{-1}x \Rightarrow dU=\frac{1}{\sqrt{1-x^2}}dx
    dV = x \cdot dx \Rightarrow V = \frac{x^2}{2}
    Then
    I=VU-\int V\cdot du=sin^{-1}x \cdot \frac{x^2}{2} -\int \frac{x^2}{2}\cdot \frac{1}{\sqrt{1-x^2}}dx

    I get stuck here, using U and V for new functions and using the relationship again doesn't lead me anywhere.
    Hi, looking at the new integral that you have to solve, you should notice the denominator suggests a possible solution using a trigonometric substitution. Consider trying:

    x = \sin \theta \Rightarrow dx = \cos \theta d\theta

    And then working out that new integral. You will get something that may require the use of an identity of the form:
    \sin^2 \theta = cos2\theta/2 + 1/2

    or something like that (ill leave the details to you).
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