Programmed tutorial: Integration of the arccos: Example 1

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Find the integral .

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What is a good method to try on a complicated integrand like arccos(3x) where we know its derivative but not its integral?

Integration by Parts frequently works. It requires doing another integral, but that second one may be simpler.

State integration by parts in general, using u and dv.

What should we take as u?

If we take u = arccos(3x), we will be able to use the derivative of the arccos in forming du.

Using a general variable, w, state the derivative of
arccos(w).

Use this to find the du we need for
our Integration by Parts.

Note the parentheses ( ) in the radical.

Complete the numerator.

State what dv is equal to in our problem.

As usual, it is equal to the rest of the problem.
Here we have dv = dx

Find v for our problem.

As always, we get v by integrating dv.

What do we get?

We get

Substitute these results into our integral.

becomes

Equation 1

Now we have the challenge of doing this new integral. What method might work here?

Since there is an x² in the denominator, and the combination x*dx in the numerator, substitution may help.

Let’s work on just the second integral, including the negative sign:

Equation 2
We’ll substitute, using a variable t = 1 – (3 x)². Find dt.

We get

Express x*dx in terms of t and dt.

Rewrite the integral in Equation 2 in terms of t and dt.

What rule of integration can we use here?

The Power Rule.

What is the power here?

The power is

Do the integral.

Express this partial result in terms of x using our definition above.

With t = 1 – (3x)²^,we get

Combine this result with our earlier results in Equation 1 to solve the original problem.

Equation 1 becomes Equation 3 result of integration of invcos by parts (cos^(-1))

How can we check this result?

We can differentiate the result,

and compare with the integrand of the original problem. What rule can we use on the first term?

The Product Rule.

Do the differentiation.

We get

Simplify this result and use a radical in the third term.

Does it check?

Since the third term reduces to be the negative of the first term, this result is the same.

Let’s try solving this same problem by using the Table of Integrals.
What is the number of the integral in the table which applies?

Number 51 applies here.

What is the value of “a” for this problem?

a = 3

What is the result of the integral using this value of “a”

Does this agree with our previous result in Equation 3?

Yes.

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