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Math Help - [SOLVED] Integral, checking a result

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Integral, checking a result

    I hope this time I didn't make an error. I also need your help for a possible little simplification.
    \int \sqrt{1-x^2}dx I call this integral \Psi \Lambda \Xi for the sake of beautiful. I made a trig-sub : Let x=\sin (\theta), dx=\cos (\theta )d\theta.
    \Rightarrow \Psi \Lambda \Xi =\int \sqrt{1-\sin ^2 (x)}\cos (\theta)d\theta=\int \sqrt{\cos ^2(\theta)}\cos (\theta)d\theta=\int \cos^2(\theta)d\theta.
    Now integration by parts : let u'=\cos (\theta), u=\sin (\theta), v=\cos (\theta), v'=-\sin (\theta) so our beautiful \Psi \Lambda \Xi is equal to \sin (\theta)\cos (\theta)+\int \sin^2(\theta)d\theta=\sin (\theta)\cos(\theta)+\int 1-\cos^2 (\theta)d\theta= \sin(\theta)\cos(\theta)+\theta-\int \cos^2(\theta)d\theta \Leftrightarrow 2\int \cos^2(\theta)d\theta= \sin(\theta)\cos(\theta)+\theta \Leftrightarrow \int \cos^2(\theta)d\theta= \frac{\sin(\theta)\cos(\theta)+\theta}{2}+C. Now I backsub and I get that \Psi \Lambda \Xi = \frac{x\cos[\arcsin(x)]+\arcsin(x)}{2}+C. Is my result correct? If no, where did I do errors? And can I simplify more this \cos[\arcsin(x)]? Thanks in advance!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by arbolis View Post
    I hope this time I didn't make an error. I also need your help for a possible little simplification.
    \int \sqrt{1-x^2}dx I call this integral \Psi \Lambda \Xi for the sake of beautiful. I made a trig-sub : Let x=\sin (\theta), dx=\cos (\theta )d\theta.
    \Rightarrow \Psi \Lambda \Xi =\int \sqrt{1-\sin ^2 (x)}\cos (\theta)d\theta=\int \sqrt{\cos ^2(\theta)}\cos (\theta)d\theta=\int \cos^2(\theta)d\theta.
    Now integration by parts : let u'=\cos (\theta), u=\sin (\theta), v=\cos (\theta), v'=-\sin (\theta) so our beautiful \Psi \Lambda \Xi is equal to \sin (\theta)\cos (\theta)+\int \sin^2(\theta)d\theta=\sin (\theta)\cos(\theta)+\int 1-\cos^2 (\theta)d\theta= \sin(\theta)\cos(\theta)+\theta-\int \cos^2(\theta)d\theta \Leftrightarrow 2\int \cos^2(\theta)d\theta= \sin(\theta)\cos(\theta)+\theta \Leftrightarrow \int \cos^2(\theta)d\theta= \frac{\sin(\theta)\cos(\theta)+\theta}{2}+C. Now I backsub and I get that \Psi \Lambda \Xi = \frac{x\cos[\arcsin(x)]+\arcsin(x)}{2}+C. Is my result correct? If no, where did I do errors? And can I simplify more this \cos[\arcsin(x)]? Thanks in advance!
    overkill.

    just recall the identity that \cos^2 \theta = \frac {1 + \cos 2 \theta}2 and integrate that. which is a lot easier (and remember to back-substitute when done and write your answer in terms of x)
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  3. #3
    MHF Contributor arbolis's Avatar
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    overkill.
    Does that mean I'm right?
    just recall the identity that and integrate that. which is a lot easier (and remember to back-substitute when done and write your answer in terms of x)
    I got that \Psi \Lambda \Xi=\frac{\arcsin (x)}{2}+\frac{\sin(2\arcsin(x))}{4}+C. Is that good?
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  4. #4
    Super Member wingless's Avatar
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    \cos \arcsin x = \sqrt{1-x^2}

    Jhevon's solution is the easiest and most common one. It's also possible to solve this without using any trigonometric substitution.

    \int \sqrt{1-x^2}~dx = \int \frac{1-x^2}{\sqrt{1-x^2}}~dx =  \arcsin x - \int \frac{x^2}{\sqrt{1-x^2}}~dx = \arcsin x + x\sqrt{1-x^2} -\int \sqrt{1-x^2}~dx

    \int \sqrt{1-x^2}~dx = \arcsin x + x\sqrt{1-x^2} -\int \sqrt{1-x^2}~dx

    \int \sqrt{1-x^2}~dx = \frac{\arcsin x + x\sqrt{1-x^2}}{2}
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