How do you integrate int (x+5)/sqrt (9-(x-3)^2) dx?

1 Answer
Aug 4, 2016

8sin^(-1)((x-3)/3) - sqrt(9-(x-3)^2) + C

Explanation:

I hope you like substitutions, because we're about to do a lot of them. To start, let u = x-3 implies du = dx

Integral becomes:

int (u+8)/(sqrt(9-u^2)) du

Now we want to deal with the denominator. This type of function will often involve a trig substitution - if you learn/can derive the pythagorean identities they are very helpful in figuring out what will cancel out etc. Whatever we choose for u in this case, it should have a coefficient of 3 so that we can take the 9 outside the square root.

Let's go with u = 3sintheta implies du = 3costheta d theta and theta = sin^(-1)(u/3)

Integral becomes:

3int (3sintheta + 8)/(3sqrt(1 - sin^2theta)) costhetad theta

1 - sin^2theta = cos^2theta so:

3int (3sintheta + 8)/(3sqrt(cos^2theta))cos thetad theta

=int(3sintheta+8)d theta

Can split this up into two seperate integrals:

=3int sintheta d theta + 8 int d theta

= -3costheta + 8theta + C

Substitute back in that theta = sin^(-1)(u/3)

= 8sin^(-1)(u/3) - 3cos(sin^(-1)(u/3)) + C

We need to figure out what the general form of cos(sin^(-1)(phi)) is:

Consider y = sin^(-1)(phi)

implies phi = sin(y)

phi^2 = sin^2(y)

phi^2 = 1 - cos^2(y)

cos(y) = cos(sin^(-1)(phi)) = sqrt(1 - phi^2)

Hence cos(sin^(-1)(u/3)) = sqrt(1 - (u/3)^2) = 1/3sqrt(9 - u^2)

Solution becomes:

8sin^(-1)(u/3) - sqrt(9-u^2) + C

Now back substitute u = x-3

Solution is:

8sin^(-1)((x-3)/3) - sqrt(9-(x-3)^2) + C