How do you integrate √(sinθ + 1)?

[Kavorka]

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.

 

[saminator910]

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

$\displaystyle\int \sqrt{1+\sin \theta} d \theta$

sub in.

$\theta = \sin^{-1}x$

$d \theta = \displaystyle\frac{1}{\sqrt{1-x^{2}}}$

$\displaystyle\int \frac{\sqrt{x+1}}{\sqrt{(1+x)(1-x)}}dx$

$\displaystyle \int \frac{1}{1-x}dx$

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

$-2\sqrt{1-\sin\theta}+c$

 

[disregardthat]

Ok, can someone confirm this for me? I must be doing something really stupid but I don't know what I'm doing wrong.

$\displaystyle\int \sqrt{1+\sin \theta} d \theta$

sub in.

$\theta = \sin^{-1}x$

$d \theta = \displaystyle\frac{1}{\sqrt{1-x^{2}}}$

$\displaystyle\int \frac{\sqrt{x+1}}{\sqrt{(1+x)(1-x)}}dx$

$\displaystyle \int \frac{1}{1-x}dx$

integrate and resub, but that doesn't really work out.

$-2\sqrt{1-\sin\theta}+c$

It's correct. The derivative of $-2\sqrt{1-\sin\theta}$ is
$$-2\frac{-\cos\theta}{2\sqrt{1-\sin\theta}} = \frac{\cos\theta}{\sqrt{1-\sin\theta}} = \frac{\sqrt{1-\sin^2\theta}}{\sqrt{1-\sin\theta}} = \sqrt{1+\sin \theta}$$

But apparently we are dividing by 0 somewhere here. The problem with the substitution is that we're restricting our domain of $\theta$ to $[-\frac{\pi}{2},\frac{\pi}{2}]$, 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 $x = \tan(\frac{\theta}{2})$.

EDIT: upon some calculations i think this might not work

 

[Saitama]

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. :tongue2: