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Integrating √(sinθ + 1)

  1. May 19, 2014 #1
    I have a polar arc length problem that comes down to integrating √(sinθ + 1). Through double u-sub and trig sub I got it to be -2√(1 - sinθ) but that seems to be wrong. Wolfram Alpha says that the integral is [2√(sinθ + 1)(sin(θ/2) - cos(θ/2)] / [sin(θ/2) + cos(θ/2). I'm wondering how this is obtained.
  2. jcsd
  3. May 19, 2014 #2
    Ok, can someone check this for me? I must be doing something stupid but I don't know what I'm doing wrong, this has to be some special case

    [itex]\displaystyle\int \sqrt{1+\sin \theta} d \theta[/itex]

    sub in.

    [itex]\theta = \sin^{-1}x[/itex]

    [itex]d \theta = \displaystyle\frac{1}{\sqrt{1-x^{2}}}[/itex]

    [itex]\displaystyle\int \frac{\sqrt{x+1}}{\sqrt{(1+x)(1-x)}}dx[/itex]

    [itex]\displaystyle \int \frac{1}{1-x}dx[/itex]

    integrate and resub, but that doesn't really work out, don't know what to tell you I got the same answer

    Last edited: May 19, 2014
  4. May 19, 2014 #3


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    It's correct. The derivative of [itex]-2\sqrt{1-\sin\theta}[/itex] is
    [tex]-2\frac{-\cos\theta}{2\sqrt{1-\sin\theta}} = \frac{\cos\theta}{\sqrt{1-\sin\theta}} = \frac{\sqrt{1-\sin^2\theta}}{\sqrt{1-\sin\theta}} = \sqrt{1+\sin \theta}[/tex]

    But apparantly we are dividing by 0 somewhere here. The problem with the substitution is that we're restricting our domain of [itex]\theta[/itex] to [itex][-\frac{\pi}{2},\frac{\pi}{2}][/itex], and x to [-1,1]. When differentiating we are also further restricted to x in (-1,1).

    What I've learned when doing integrals involving trigonometry is this: try the magical substitution [itex]x = \tan(\frac{\theta}{2})[/itex].

    EDIT: upon some calculations i think this might not work
    Last edited: May 19, 2014
  5. May 19, 2014 #4
    Notice that:
    $$\sqrt{1+\sin x}=\sin\left(\frac{x}{2}\right)+\cos\left(\frac{x}{2}\right)$$
    Integrating gives:
    $$\int \sqrt{1+\sin x}\,dx=2\left(\cos\left(\frac{x}{2}\right)-\sin\left(\frac{x}{2}\right)\right)+C$$
    Multiply and divide by ##\sin\left(\frac{x}{2}\right)+\cos\left(\frac{x}{2}\right)## i.e
    $$2\left(\cos\left(\frac{x}{2}\right)-\sin\left(\frac{x}{2}\right)\right)=\frac{2\left(\cos\left(\frac{x}{2} \right) - \sin\left(\frac{x}{2}\right)\right)\left( \sin\left(\frac{x}{2}\right)+\cos \left (\frac{x}{2}\right)\right)}{\sin\left(\frac{x}{2}\right)+\cos \left (\frac{x}{2}\right)}$$
    $$\sin\left(\frac{x}{2}\right)+\cos\left(\frac{x}{2}\right)=\sqrt{1+ \sin x}$$
    $$\int \sqrt{1+\sin x}\,dx=\frac{2\sqrt{1+\sin x}\left(\cos\left(\frac{x}{2}\right)-\sin\left(\frac{x}{2}\right)\right)}{\sin\left(\frac{x}{2}\right)+\cos \left (\frac{x}{2}\right)}+C$$
    The above result is same as ##2\sqrt{1-\sin x}+C##, W|A likes to complicate the things. :tongue2:
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