Integrating √(1+sin x) - Calculus II

[Bachelier]

How would you integrate this. I tried all integration methods available in Calculus II to no avail.

∫√(1+sin x) dx

 

[Bohrok]

Did you try u = sinx? That should give you something easy to work with.

 

[Bachelier]

Did you try u = sinx? That should give you something easy to work with.

I tried but d(sin x)/dx is cos x. We're missing the cos(x) here. :)

 

[lanedance]

is this indefinite or does it have limits? if there is limits you may be able to use the symmetry with cos to simplify things

 

[Bachelier]

is this indefinite or does it have limits? if there is limits you may be able to use the symmetry with cos to simplify things

It has limits [0,∏]. I know the cosine is a symmetrical function, but could you please expand a little on your symmetry argument in solving this integral.
thank you

 

[lanedance]

well if you draw the sin and cos functions, you might be able to convince yourself that (could also do it by hand using pi/2 funtion shifts & negative angles):

based on symmetry of the sin function around pi/2
\int_0^{\pi} dx \frac{1}{\sqrt{1+sin(x)}} = 2 \int_0^{\pi/2} dx \frac{1}{\sqrt{1+sin(x)}}

and based on symmetry of the sin & cos functions on the interval [0,pi/2]
\int_0^{\pi/2} dx \frac{1}{\sqrt{1+sin(x)}} = \int_0^{\pi/2} dx \frac{1}{\sqrt{1+cos(x)}}

so you could re-write the integral as
\int_0^{\pi/2} dx (\frac{1}{\sqrt{1+sin(x)}} + \frac{1}{\sqrt{1+cos(x)}})

unfortunately after all that, I'm not sure whether it helps, remember a similar looking integral and though it might be worth a go... sorry ;(

 

[lanedance]

so if you bought up to here
\int_0^{\pi} dx \frac{1}{\sqrt{1+sin(x)}} = 2 \int_0^{\pi/2} dx \frac{1}{\sqrt{1+cos(x)}}

how about using half angle
cos(x) = 2cos^2(x/2)-1

then
\int_0^{\pi/2} dx \frac{1}{\sqrt{1+cos(x)}}= \int_0^{\pi/2} dx \frac{1}{\sqrt{2} cos(x/2)}

which is getting maybe a little more tractable...?

 

[Bohrok]

I hadn't looked at it too closely at first, but this one is a little tricky where you need another subtle substitution:

\int \sqrt{1 + sinx} ~dx
u = sinx → x = sin^-1u

du = cosx dx

\frac{du}{\cos(\sin^{-1}u)} = dx

\int \frac{\sqrt{1 + u}}{\cos(\sin^{-1}u)} du