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Integrating √(sinθ + 1)

by Kavorka
Tags: √sinθ, integrating
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Kavorka
#1
May19-14, 04:01 PM
P: 33
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.
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saminator910
#2
May19-14, 04:46 PM
P: 94
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

[itex]-2\sqrt{1-\sin\theta}+c[/itex]
disregardthat
#3
May19-14, 05:05 PM
Sci Advisor
P: 1,742
Quote Quote by saminator910 View Post
Ok, can someone confirm this for me? I must be doing something really stupid but I don't know what I'm doing wrong.

[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.

[itex]-2\sqrt{1-\sin\theta}+c[/itex]
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

Pranav-Arora
#4
May19-14, 11:07 PM
P: 3,806
Integrating √(sinθ + 1)

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
$$2\left(\cos\left(\frac{x}{2}\right)-\sin\left(\frac{x}{2}\right)\right)=\frac{2\left(\cos\left(\frac{x}{2} \right) - \sin\left(\frac{x}{2}\right)\right)\left( \sin\left(\frac{x}{2}\right)+\cos \left (\frac{x}{2}\right)\right)}{\sin\left(\frac{x}{2}\right)+\cos \left (\frac{x}{2}\right)}$$
Rewrite
$$\sin\left(\frac{x}{2}\right)+\cos\left(\frac{x}{2}\right)=\sqrt{1+ \sin x}$$
Hence,
$$\int \sqrt{1+\sin x}\,dx=\frac{2\sqrt{1+\sin x}\left(\cos\left(\frac{x}{2}\right)-\sin\left(\frac{x}{2}\right)\right)}{\sin\left(\frac{x}{2}\right)+\cos \left (\frac{x}{2}\right)}+C$$
The above result is same as ##2\sqrt{1-\sin x}+C##, W|A likes to complicate the things.


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