# Hard Integration

1. Sep 27, 2009

### Bachelier

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

∫√(1+sin x) dx

2. Sep 27, 2009

### Bohrok

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

3. Sep 28, 2009

### Bachelier

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

4. Sep 28, 2009

### 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

5. Sep 28, 2009

### Bachelier

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

6. Sep 28, 2009

### 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 ;(

Last edited: Sep 28, 2009
7. Sep 28, 2009

### lanedance

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

$$cos(x) = 2cos^2(x/2)-1$$

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...?

8. Sep 28, 2009

### 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

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

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