Question #e9b24

Question

882 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-16T10:54:59

$int\ dx/sqrt(-x^2-8x)=arcsin((x+4)/4)+C$

$\ \ \ \ \ \ int\ dx/sqrt(-x^2-8x)$

Completing the square, we have
$=int\ dx/sqrt(16-(x+4)^2)$

Substitute $u=x+4$ and $du=dx$ :
$=int\ (du)/sqrt(16-u^2)$

Substitute $u=4sin(theta)$ and $du=4cos(theta)\ d theta$ :
$=int\ (4cos(theta))/sqrt(16-16sin^2(theta))\ d theta$

$=int\ (4cos(theta))/(4cos(theta))\ d theta$

$=int\ d theta$

$=theta+C$

$=arcsin(u/4)+C$

$=arcsin((x+4)/4)+C$