Question #bb4fb

1 Answer
Dec 15, 2017

1/2arcsin((2x)/3)+c

Explanation:

int1/sqrt(9-4x^2)dx

it can be rewritten

int1/sqrt(3^2-(2x)^2)dx

trigonometrical substitution

2x=3sin(theta)

dx=(3cos(theta)d(theta))/2

sqrt(9-4x^2)=3cos(theta)

in the integral

int1/sqrt(3^2-(2x)^2)dx=int(3cos(theta)d(theta))/(2*3cos(theta)

int(3cos(theta)d(theta))/(2*3cos(theta)=int(d(theta))/2

1/2int(d(theta))=1/2(theta)+c

but
theta=arcsin((2x)/3)

1/2int(d(theta))=1/2arcsin((2x)/3+c)

int1/sqrt(9-4x^2)dx=1/2arcsin((2x)/3)+c