Hello, I have a question concerning trigonometric substitution.

let's say we have

integral of dx/sqrt(9x^2 + 4),

so after doing a few steps we have:

2/3 integral of sec0/sqrt((2tan0)^2 + 4*) d0 (the * is for later on)

the next step turns into:

2/3 integral of sec0/(sqrt(4tan^2 0 + 1))

I know that the sqrt(4 tan^2 0 + 1) is because of the power of 2 from the first step, but where does the other + 4 disappear (the one I put a * next to).

You need 3x = 2tanθ, so

9x^2 = 4tan^2 θ
9x^2+4 = 4tan^2 θ + 4 = 4sec^2 θ

thank you Steve

To answer your question, let's go back to the step where we have:

2/3 integral of sec0/sqrt((2tan0)^2 + 4*) d0

The * indicates a change that will be made in the next step. At this point, we want to rewrite the expression inside the square root in terms of the variable tan0. Looking at the expression (2tan0)^2, we notice that it is the square of (2tan0), which simplifies to 4(tan0)^2.

Now, let's focus on the remaining + 4. We want to make a substitution, so we manipulate the expression inside the square root. By adding and subtracting 4(tan0)^2, we can rewrite it as:

(4(tan0)^2 + 4 - 4(tan0)^2)

The purpose of introducing 4 and subtracting 4(tan0)^2 is to create a perfect square trinomial, which allows us to rewrite it using a popular trigonometric identity.

Now, let's simplify the expression further:

(4(tan0)^2 + 4 - 4(tan0)^2)

= 4 - 4(tan0)^2 + 4(tan0)^2

= 4

Thus, we see that the +4 term effectively disappears since it cancels out with -4(tan0)^2. Therefore, we are left with:

2/3 integral of sec0/sqrt(4tan^2 0 + 1)

I hope this explanation helps clarify why the +4 term disappears in the subsequent step.