Trigonometric Substitution: int sqrt(4-x^2) dx

[hank]

Ok, here's the problem:

S sqrt(4-x^2) dx (1)

[mathy stuff here]

= S sqrt (cos^2 (a)) cos(a) da
= S cos^2(a) da (2)

= a/2 + 1/4 sin(2a) + C

= 1/2 sqrt(4-x^2) + 1/4 sin(2(arctan(x/2))) + C (3)

= 1/2 sqrt (4 - x^2) + 2arctan (x/2) + C (4)

Am I correct in going from (1) to (2)?

Is (3) to (4) correct? If so, how does one get to there?
I came up with (3) on my own, but the book shows (4) as the answer, and I'm not sure how to transform sin(2(arctan(x/2))) into 2arctan (x/2).

 

[galactus]

\(\displaystyle \L\\\int{\sqrt{4-x^{2}}}dx\)

As you know, \(\displaystyle x=2sin(t), \;\ dx=2cos(t)dt\)

I will skip ahead to:

\(\displaystyle \L\\4\int{cos^{2}(t)}dt=2(sin(t)cos(t)+t)\)

Using the triangle, make the appropriate subs to get back in terms of x.

\(\displaystyle \L\\sin(t)=\frac{x}{2}, \;\ cos(t)=\frac{\sqrt{4-x^{2}}}{2}, \;\ t=sin^{-1}(\frac{x}{2})\)

Then we have:

\(\displaystyle \L\\2\left(\frac{x}{2}\cdot\frac{4-x^{2}}{2}+sin^{-1}(\frac{x}{2})\right)\)

\(\displaystyle =\L\\2sin^{-1}(\frac{x}{2})+\frac{x\sqrt{4-x^{2}}}{2}+C\)

I don't know where the arctan is coming from, but for your info, \(\displaystyle \L\\sin(2tan^{-1}(\frac{x}{2}))=\frac{4x}{x^{2}+4}\)