Indefinite integral

Question

Evaluate the following:

∫9/(√4-x2) dx

Answer 1

You need to hire one of hese tutors doing all your work for free.  Not very kosher of you.

Answer 2

First pull the constant 9 out of the numerator.

 

9∫1/√(4-x^2) dx

 

This one requires some recognition. The integrals of inverse trig functions take forms similar to the one shown. In this case, the trig function we're looking at is arcsin, since

 

∫1/√(1-z^2) dz = arcsin(z) + C

 

The only issue is that you need to have 1-z^2 in the square root, not 4-x^2. I put 'z' instead of 'x' because the z term is allowed to have constants attached to it. You'll see in a second what I mean.

 

Factor out the 4 in the square root.

 

√(4-x^2) = √(4(1-(1/4)*x^2))

 

√4 is a nice and simple number, which gives us a constant we can again pull out of the integral. Since it's in the denominator, it would go under the 9 currently outside the integral. You're then left with:

9/2 * ∫1/√(1-(1/4)*x^2) dx

 

We just need one more step to make the term in the square root look like 1-z^2. Place the 1/4 scalar into the squared function with x. Fortunately, the arithmetic will look really clean.

 

9/2 * ∫1/√(1-(1/2*x)^2) dx

 

So in this case, z = 1/2*x

 

∫1/√(1-z^2) dz = arcsin(z) + C

 

So, substituting your relevant variables:

 

9/2 * ∫1/√(1-(1/2*x)^2) dx = 9/2 * arcsin(1/2*x) + C