I hope you like substitutions, because we're about to do a lot of them. To start, let u = x-3 implies du = dxu=x−3⇒du=dx
Integral becomes:
int (u+8)/(sqrt(9-u^2)) du∫u+8√9−u2du
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 uu 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)u=3sinθ⇒du=3cosθdθandθ=sin−1(u3)
Integral becomes:
3int (3sintheta + 8)/(3sqrt(1 - sin^2theta)) costhetad theta3∫3sinθ+83√1−sin2θcosθdθ
1 - sin^2theta = cos^2theta1−sin2θ=cos2θ so:
3int (3sintheta + 8)/(3sqrt(cos^2theta))cos thetad theta3∫3sinθ+83√cos2θcosθdθ
=int(3sintheta+8)d theta=∫(3sinθ+8)dθ
Can split this up into two seperate integrals:
=3int sintheta d theta + 8 int d theta=3∫sinθdθ+8∫dθ
= -3costheta + 8theta + C=−3cosθ+8θ+C
Substitute back in that theta = sin^(-1)(u/3)θ=sin−1(u3)
= 8sin^(-1)(u/3) - 3cos(sin^(-1)(u/3)) + C=8sin−1(u3)−3cos(sin−1(u3))+C
We need to figure out what the general form of cos(sin^(-1)(phi))cos(sin−1(ϕ)) is:
Consider y = sin^(-1)(phi)y=sin−1(ϕ)
implies phi = sin(y)⇒ϕ=sin(y)
phi^2 = sin^2(y)ϕ2=sin2(y)
phi^2 = 1 - cos^2(y)ϕ2=1−cos2(y)
cos(y) = cos(sin^(-1)(phi)) = sqrt(1 - phi^2)cos(y)=cos(sin−1(ϕ))=√1−ϕ2
Hence cos(sin^(-1)(u/3)) = sqrt(1 - (u/3)^2) = 1/3sqrt(9 - u^2)cos(sin−1(u3))=√1−(u3)2=13√9−u2
Solution becomes:
8sin^(-1)(u/3) - sqrt(9-u^2) + C8sin−1(u3)−√9−u2+C
Now back substitute u = x-3u=x−3
Solution is:
8sin^(-1)((x-3)/3) - sqrt(9-(x-3)^2) + C8sin−1(x−33)−√9−(x−3)2+C
