FORM a ² + x ²

Let x = a tan q, then dx = a sec ² q dq. Use this substitution to simplify a ² + x ² to a single term like this.

a ² + x ² = a ² + a ² tan ² q = a ²(1 + tan ² q ) = a ²sec ² q

Here is the reference triangle for this substitution.

FORM a ² - x ²

Let x = a sin q, then dx = a cos q dq. Use this substitution to simplify a ² - x ² to a single term like this.

a ² - x ² = a ² - a ²sin ² q = a ²(1 - sin ² q ) = a ²cos ² q

Here is the reference triangle for this substitution.

FORM x ² - a ²

Let x = a sec q, then dx = a sec q tan q dq. Use this substitution to simplify x ² - a ² to a single term like this.

x ² - a ² = a ²sec ² q - a ² = a ²(sec ² q - 1) = a ² tan ² q

Here is the reference triangle for this substitution.

You should notice that a = 2, so the expression under the radical sign is of the form a ² - x ². Therefore, let x = 2sin q, then dx = 2cos q dq, x ² = 4sin ² q, and 4 - x ² = 4 - 4sin ² q = 4(1 - sin ² q) = 4cos ² q.

Now let us simplify the integrand.

Now remember that cot ² q = csc ² q -1, so the integral will now become the following.

Now we have to replace the cot q and q with their equivalent expression that are in terms of the original variable. First, let us solve for q in terms of x.

Now, using what q is, let us construct the reference triangle for this substitution. Remember that the sine is defined to be opposite over hypotenuse. From the reference triangle we can determine the expression for the cot q.

Let u = 2x and a = 3, therefore this integrand is of the form u ² - a ².

 Now to substitute this result in for the original integrand.

Let u = sin q, then du = cos q dq. 

We must now construct the reference triangle for this substitution. First of all we must solve 2x = 3sec q for sec q.

Now we will use this to design the reference triangle. Remember that the secant is defined to be hypotenuse over adjacent. Here is the reference triangle, and from this triangle, we can determine the value for the csc q.

Now, we will substitute the value for the csc q in to complete the problem.

The first thing that you should notice that x ² + 4 is of the form u ² + a ², so we will let x = 2tan q, and dx = 2sec ² q. Also, x ² + 4 = 4tan ² q + 4 = 4(tan ² q + 1) = 4sec ² q.

Now we have to substitute in for csc q and cot q. To do this, we must construct the reference triangle for x = 2 tan q, so let us solve for tan q first.

Now using what tan q equals and the fact that tangent is defined to be opposite over adjacent, we will construct the triangle.

Now substitute these values in.

Let us first complete the square for 5 - 4x - x ².

Therefore,

Notice that the integrand is of the form a ²- u ², so let x + 2 = 3sin q, then dx = 3cos q dq, and 9 - (x + 2) ² = 9 - 9sin ² q = 9(1 - sin ² q) = 9cos ² q. Now to convert the integrand.

Making this substitution, the integral now becomes the following.

The reference triangle for x + 2 = 3sin q is the following.

Notice that this integrand is in the form a ² + u ², so let x = 4tan q, then dx = 4sec ² q, and 16 + x ² = 16 + 16tan ² q = 16(1 + tan ² q) = 16sec ² q.

What will x ⁴ be equal to? x ⁴ = (4tan q) ⁴ = 256tan ⁴ q.

Let u = csc q, then du = -cot q csc q dq.

 Here is the reference triangle for x = 4tan q.

Therefore,

The integrand is of the form u ² - a ², so let u = 2sec q , then du = 2sec q tan q dq and

u ² - 4 = 4sec ² q - 4 = 4tan ² q.

Let p = tan q, then dp = sec ² q dq.

 Now to construct the reference triangle for u = 2sec q .

 Now substitute in for u and we are done with this integral.

First of all, I am going to multiply the integrand by a form of one. The reason why I want to do this is so I can let u = e ^x.

Now, let u = e ^x, then du = e ^x dx. When x = 0, u = 1, and when x = 1, u = e.

Notice that the integrand is of the form a ² - u ², so let u = 4sin q , then du = 4cos q . Also, 16 - u ² = 16 - 16sin ² q = 16(1 - sin ² q ) = 16cos ² q .

Now find the reference triangle for u = 4sin q .

We also need to solve for q .