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

Question

Aurora Alvarado · Accepted Answer

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

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#

Isaiah Allen · Answer

undefined To integrate $ \frac{{\text{int}(x+5)}}{{\sqrt{9-(x-3)^2}}} $, you can use the trigonometric substitution method. Let $ x - 3 = 3 \sin(\theta) $, then $ dx = 3 \cos(\theta) d\theta $. After substitution and simplification, the integral becomes $ \int \frac{{8 + 3 \sin(\theta)}}{{3 \cos(\theta)}} d\theta $. This simplifies to $ \int (8\sec(\theta) + 3\tan(\theta)) d\theta $. Finally, integrate each term separately to get $ 8\ln|\sec(\theta) + \tan(\theta)| + \frac{3}{2} \ln|3 \sin(\theta) + 3 \cos(\theta)| + C $, where $ C $ is the constant of integration.