1. ## integration problem

prove that ... integrate sqrt [(1-x)/(1+x)] dx ( upper limit = 1 , lower limit =0 ) = pi/2 - 1

by using x = cos 2 theta

2. Originally Posted by ose90
prove that ... integrate sqrt [(1-x)/(1+x)] dx ( upper limit = 1 , lower limit =0 ) = pi/2 - 1

by using x = cos 2 theta

Note:

1. $\frac{1 - \cos (2 \theta)}{1 + \cos (2 \theta)} = \frac{1 - \cos^2 \theta + \sin^2 \theta}{1 + \cos^2 \theta - \sin^2 \theta} = \frac{2 \sin^2 \theta}{2 \cos^2 \theta} = \tan^2 \theta$.

2. $dx = -2 \sin (2 \theta) \, d \theta = -4 \sin \theta \cos \theta \, d \theta$.

3. Originally Posted by mr fantastic
Note:

1. $\frac{1 - \cos (2 \theta)}{1 + \cos (2 \theta)} = \frac{1 - \cos^2 \theta + \sin^2 \theta}{1 + \cos^2 \theta - \sin^2 \theta} = \frac{2 \sin^2 \theta}{2 \cos^2 \theta} = \tan^2 \theta$.

2. $dx = -2 \sin (2 \theta) \, d \theta = -4 \sin \theta \cos \theta \, d \theta$.

Thanks you very much, I wrongly conceive that 1-cos 2 theta = sin theta and 1+ cos 2 theta = cos theta , no wonder I couldn't solve it!

Now I can solve it, thank you!

4. Originally Posted by mr fantastic
Note:

1. $\frac{1 - \cos (2 \theta)}{1 + \cos (2 \theta)} = \frac{1 - \cos^2 \theta + \sin^2 \theta}{1 + \cos^2 \theta - \sin^2 \theta} = \frac{2 \sin^2 \theta}{2 \cos^2 \theta} = \tan^2 \theta$.

2. $dx = -2 \sin (2 \theta) \, d \theta = -4 \sin \theta \cos \theta \, d \theta$.

Hmm, I have solved it ! Thanks if you have come here again.