How do you use substitution to integrate $sqrt(4-x^2) dx$ ?

Question

35617 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-23T21:02:16

This can be done from trig substitution. Notice how

$sqrt(a^2 - x^2) prop sqrt(a^2 - a^2sin^2theta) prop sqrt(4 - x^2)$
where $a = 2$

so let:
$x = 2sintheta$
$dx = 2costhetad theta$
$sqrt(4-x^2) = 2costheta$

$=> int 2costheta*2costhetad theta$

$= 4int cos^2thetad theta$

Now you can use the identity:
$cos^2theta = (1+cos(2theta))/2$

Thus:
$= 2int d theta + 2int cos(2theta)d theta$

$= 2int d theta + 2*1/2int 2cos(2theta)d theta$

$= 2theta + sin(2theta) + C$

Since $x = 2sintheta$ , $theta = arcsin(x/2)$ .
Since $sin(2theta) = 2sinthetacostheta$ :

$sin(2theta) = (xsqrt(4-x^2))/2$

$=> color(blue)(2arcsin(x/2) + (xsqrt(4-x^2))/2 + C)$

How do you use substitution to integrate sqrt(4-x^2) dxsqrt(4-x^2) dx?