More Integration by Substitution (1 Viewer)

Crosswinds · Jan 17, 2010

Could someone please show me why the answer to this:

∫ (sinx)/(√(1 + 2cosx)) dx {with limits pi/2 and 0} using the substitution y = 1 + 2cosx

is

√3 - 1?

I keep getting a quarter of this answer??

Thanks! (Sorry I don't know how to make my limits pretty)

Cazic · Jan 17, 2010

Letting y = 1 + 2cos(x) we have -dy/2 = sin(x) dx and:

$\int \frac{\sin(x)}{\sqrt{1 + 2\cos(x)} }\, dx = - \int \frac{dy}{2\sqrt{y}} = -\sqrt{y} + C = -\sqrt{1 + 2\cos(x)} + C.$

So

$\int_{0}^{\frac{\pi}{2}} \frac{\sin(x)}{\sqrt{1 + 2\cos(x)}} \, dx = -\sqrt{1 + 2\cos(x)} \ \biggr|_{0}^{\frac{\pi}{2}} = -1 + \sqrt{3}.$

Edit: I must have missed a minus sign somewhere if your answer is correct /shrug
Edit2: Found it

jet · Jan 17, 2010

Don't forget the dy in the second integrand

gurmies · Jan 17, 2010

Cazic · Jan 17, 2010

jetblack2007 said:
Don't forget the dy in the second integrand

Muchos gracias senior.

Crosswinds · Jan 18, 2010

Ah yeahhhh oooops I just did something really dumb. Thanks for your help!

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