Results 1 to 5 of 5

Math Help - Confusing integral

  1. #1
    VkL
    VkL is offline
    Member
    Joined
    Oct 2008
    Posts
    96

    Confusing integral

    How do I start this?


    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Abu-Khalil's Avatar
    Joined
    Oct 2008
    From
    Santiago
    Posts
    148
    Let t=\sin^{-1}x\Rightarrow dt = \frac{dx}{\sqrt{1-x^2}} so \int \frac{xe^{\sin^{-1}x}}{\sqrt{1-x^2}}dx=\int \sin x e^x dx=I and I=e^x\sin x-\int \cos x e^xdx = e^x \sin x - (\cos xe^x + \int \sin x e^x dx) =e^x(\sin x - \cos x)-I\Rightarrow I =e^x\frac{\sin x -\cos x }{2}+C.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    VkL
    VkL is offline
    Member
    Joined
    Oct 2008
    Posts
    96
    ^ I lost you at I ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member Abu-Khalil's Avatar
    Joined
    Oct 2008
    From
    Santiago
    Posts
    148
    I=e^x\sin x -e^x \cos x -I \Rightarrow 2I=e^x\sin x - e^x \cos x
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,912
    Thanks
    775
    Hello, VkL!

    Abu-Khalil is correct . . . (he could have been clearer)


    20)\;\;\int\frac{xe^{\sin^{-1}\!x}}{\sqrt{1-x^2}}\,dx

    We have: . \int x\cdot e^{\sin^{1}x}\cdot\frac{dx}{\sqrt{1-x^2}}

    Let: . t \:=\: \sin^{-1}\!x \quad\Rightarrow\quad dt \:=\:\frac{dx}{\sqrt{1-x^2}}
    Note that: . x \:=\:\sin t

    Substitute: . \int \sin t\cdot e^t\cdot dt


    We will integrate by parts: . I \;=\;\int e^t\sin t\,dt

    . . \begin{array}{ccccccc}u &=& \sin t & & dv &=& e^t\,dt \\ du &=& \cos t\,dt & & v &=& e^t\end{array}

    We have: . I \;=\;e^t\sin t - \int e^t\cos t\,dt

    . . By parts: . \begin{array}{ccccccc}u &=& \cos t & & dv&=& e^t\,dt \\ du &=& -\sin t\,dt & & v &=& e^t \end{array}

    And we have: . I \;=\;e^t\sin t - \bigg[e^t\cos t + \int e^t\sin t\,dt\bigg]

    . . . . . . . . . I \;=\;e^t\sin t - e^t\cos t - \underbrace{\int e^t\sin t\,dt}_{\text{This is }I} + C

    So we have: . I \;=\;e^t\sin t - e^t\cos t - I + C

    . . . . . . . . . 2I \;=\;e^t\sin t - e^t\cos t + C

    . . . . . . . . . . I \;=\;\tfrac{1}{2}\left(e^t\sin t - e^t\cos t\right) + C

    Hence: . \int e^t\sin t\,dt \;=\;\tfrac{1}{2}\,e^t(\sin t - \cos t) + C


    Back-substitute: . t \,=\,\sin^{-1}\!x

    We have: . \frac{1}{2}e^{\sin^{-1}\!x}\bigg[\sin\left(\sin^{-1}\!{x}\right) - \cos\left(\sin^{-1}\!x\right)\bigg] + C

    . . . . . . . = \;\frac{1}{2}\,e^{\sin^{-1}\!x}\bigg[x - \sqrt{1-x^2}\bigg] + C

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Confusing Integral!
    Posted in the Calculus Forum
    Replies: 8
    Last Post: February 6th 2011, 11:10 AM
  2. Confusing integral question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 29th 2010, 07:17 AM
  3. Confusing integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 20th 2010, 10:34 AM
  4. confusing integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 15th 2010, 03:44 AM
  5. Confusing Integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 1st 2009, 02:41 AM

Search Tags


/mathhelpforum @mathhelpforum