Section 9.4 Integration with Trigonometric Functions

Subsection 9.4.1 Preview Activity

Question 9.4.1.

We already know we can do the following problem using u-substitution, with \(u = \sin(-15 x )\)

\(\displaystyle \int \sin^{4}\mathopen{}\left(-15x\right)\cos\mathopen{}\left(-15x\right) \,dx \ = \ \)

We see that \(\displaystyle \int \sin^{4} (-15 x) \cos^3(-15 x) \,dx \ \) is solved similarly, because

the exponent of \(\cos\) is odd.
the exponent of \(\sin\) is odd.
None of the above

So we’ll still use u-substitution with \(u = \sin(-15 x )\text{,}\) after first rewriting \(\cos^3 (-15 x) = \cos^2(-15 x) \cos(-15 x )\) and using a trig identity.

Select the trigonometric identity needed to solve this problem.

\(\displaystyle \displaystyle \cos^2 u = \frac{1+\cos(2 u)}{2}\)
\(\displaystyle \displaystyle 1+ \cot^2 u = \csc^2 u\)
\(\displaystyle \displaystyle \tan^2 u +1 = \sec^2 u\)
\(\displaystyle \displaystyle \sin^2 u = \frac{1-\cos(2 u)}{2}\)
\(\displaystyle \displaystyle \sin^2 u + \cos^2 u = 1\)
None of the above

Finally, put it all together to calculate \(\displaystyle \int \sin^{4}\mathopen{}\left(-15x\right)\cos^{3}\mathopen{}\left(-15x\right) \,dx \ = \ \)

Hint.

The integral involves \(\sin u\) and \(\cos u\) where the exponent of \(\cos u\) is an odd, positive integer, namely, 3. So we factor out a copy of \(\cos(-15 x)\) and use the identity \(\sin^2 u + \cos^2 u = 1\) to turn the remaining \(\cos^2(-15 x)\) into an expression involving \(\sin^2(-15 x)\text{.}\)

Thus the integral can be rewritten as

\begin{equation*} \int \sin^{4}\mathopen{}\left(-15x\right)\cos^{3}\mathopen{}\left(-15x\right) \, dx = \int \sin^{4} (-15 x) \cos^2(-15 x) \cos(-15 x) \, dx \end{equation*}

\begin{equation*} = \int \sin^{4} (-15 x) \bigg(1-\sin^2(-15 x)\bigg) \cos(-15 x) \, dx \end{equation*}

and we use u-substitution.

Letting \(u= \sin (-15 x)\) so \(du = -15 \cos (-15 x) \, dx\) and thus \(\displaystyle dx = \frac{ du}{-15 \cos (-15 x)}\text{.}\)

This turns the integral into

\begin{equation*} \int u^{4} (1-u^2) \cos(-15 x) \, \frac{ du}{-15 \cos (-15 x)} = \frac{1}{-15}\int u^{4} (1-u^2) \, du \end{equation*}

which can be multiplied out, and then each part integrated by power rule.

Question 9.4.2.

Consider the integral \(\displaystyle \int 3\sin^{2}\mathopen{}\left(9x\right) \,dx\text{.}\)

In order to integrate, the key feature is that

the exponent of \(\sin\) is odd.
the exponents of \(\sin\) and \(\cos\) are both even (remember that 0 is even)
the exponent of \(\cos\) is odd.
None of the above

Then the trigonometric identity needed to solve this problem is

\(\displaystyle \displaystyle 1+ \cot^2 u = \csc^2 u\)
\(\displaystyle \displaystyle \tan^2 u +1 = \sec^2 u\)
\(\displaystyle \displaystyle \cos^2 u = \frac{1+\cos(2 u)}{2}\)
\(\displaystyle \displaystyle \sin^2 u + \cos^2 u = 1\)
\(\displaystyle \displaystyle \sin^2 u = \frac{1-\cos(2 u)}{2}\)
None of the above

And we use this identity to rewrite the integral as \(\displaystyle \int 3\sin^{2}\mathopen{}\left(9x\right) \, dx = \ \int\) \(dx\)

And this can be split into two basic integrals and integrated, though I’m not asking for you to do that here.

Question 9.4.3.

Select all the key features we could look for to evaluate trig integrals

the exponents of both \(\sin\) and \(\cos\) are even.
the exponent of \(\tan\) is even.
the exponent of \(\sin\) is even.
the exponent of \(\tan\) is odd.
the exponent of \(\sec\) is odd.
the exponent of \(\cos\) is odd.
the exponent of \(\cos\) is even.
the exponent of \(\sec\) is even.
the exponent of \(\sin\) is odd.
None of the above
All of the above

You should be sure to have this information written down so that you can use it to solve problems in class.

Question 9.4.4.

Consider the integral \(\displaystyle \int 7\tan^{3}\mathopen{}\left(-6x\right)\sec^{3}\mathopen{}\left(-6x\right) \,dx\text{.}\)

In order to integrate, the key feature is that

the exponent of \(\sec\) is even.
the exponent of \(\tan\) is odd.
the exponent of \(\tan\) is even.
the exponent of \(\sec\) is odd.
None of the above

Then the trigonometric identity needed to solve this problem is

\(\displaystyle \displaystyle \cos^2 u = \frac{1+\cos(2 u)}{2}\)
\(\displaystyle \displaystyle \sin^2 u = \frac{1-\cos(2 u)}{2}\)
\(\displaystyle \displaystyle \tan^2 u +1 = \sec^2 u\)
\(\displaystyle \displaystyle \sin^2 u + \cos^2 u = 1\)
\(\displaystyle \displaystyle 1+ \cot^2 u = \csc^2 u\)
None of the above

So to integrate, we would use u-substitution with \(u= \sec(-6 x)\text{.}\) But first we’d have to rewrite the integral in order to have a copy of \(u\)’s derivative \(\sec(-6 x) \tan(-6 x)\text{,}\) and use a trigonometric identity to turn the remaining even powers of \(\tan(-6 x)\) into an expression of \(\sec^2(-6 x)\text{.}\)

Question 9.4.5.

Consider the integral \(\displaystyle \int 12\tan^{4}\mathopen{}\left(7x\right)\sec^{4}\mathopen{}\left(7x\right) \,dx\text{.}\)

In order to integrate, the key feature is that

the exponent of \(\tan\) is even.
the exponent of \(\sec\) is even.
the exponent of \(\sec\) is odd.
the exponent of \(\tan\) is odd.
None of the above

Then the trigonometric identity needed to solve this problem is

\(\displaystyle \displaystyle 1+ \cot^2 u = \csc^2 u\)
\(\displaystyle \displaystyle \tan^2 u +1 = \sec^2 u\)
\(\displaystyle \displaystyle \sin^2 u = \frac{1-\cos(2 u)}{2}\)
\(\displaystyle \displaystyle \cos^2 u = \frac{1+\cos(2 u)}{2}\)
\(\displaystyle \displaystyle \sin^2 u + \cos^2 u = 1\)
None of the above

So to integrate, we would use u-substitution with \(u= \tan(7 x)\text{.}\) But first we’d have to rewrite the integral in order to have a copy of \(u\)’s derivative \(\sec^2(7 x)\text{,}\) and use a trigonometric identity to turn the remaining even powers of \(\sec(7 x)\) into an expression involving \(\tan^2(7 x)\text{.}\)

Exercises 9.4.2 Exercises

1.

Evaluate the indefinite integral.

\(\displaystyle \int 5\tan^{3}\mathopen{}\left(9x\right)\sec^{3}\mathopen{}\left(9x\right) \,dx \ = \ \)

Select the trigonometric identity you used in solving this problem.

\(\displaystyle \displaystyle \tan^2 u +1 = \sec^2 u\)
\(\displaystyle \displaystyle 1+ \cot^2 u = \csc^2 u\)
\(\displaystyle \displaystyle \sin^2 u + \cos^2 u = 1\)
\(\displaystyle \displaystyle \sin^2 u = \frac{1-\cos(2 u)}{2}\)
\(\displaystyle \displaystyle \cos^2 u = \frac{1+\cos(2 u)}{2}\)
None of the above

2.

Evaluate the indefinite integral.

\(\displaystyle \int 10\tan^{4}\mathopen{}\left(-11x\right)\sec^{4}\mathopen{}\left(-11x\right) \,dx \ = \ \)

Select the trigonometric identity you used in solving this problem.

\(\displaystyle \displaystyle 1+ \cot^2 u = \csc^2 u\)
\(\displaystyle \displaystyle \sin^2 u + \cos^2 u = 1\)
\(\displaystyle \displaystyle \sin^2 u = \frac{1-\cos(2 u)}{2}\)
\(\displaystyle \displaystyle \cos^2 u = \frac{1+\cos(2 u)}{2}\)
\(\displaystyle \displaystyle \tan^2 u +1 = \sec^2 u\)
None of the above

3.

Evaluate the indefinite integral.

\(\displaystyle \int 4\cot^{3}\mathopen{}\left(12x\right)\csc^{3}\mathopen{}\left(12x\right) \,dx \ = \ \)

Select the trigonometric identity you used in solving this problem.

\(\displaystyle \displaystyle \cos^2 u = \frac{1+\cos(2 u)}{2}\)
\(\displaystyle \displaystyle \sin^2 u = \frac{1-\cos(2 u)}{2}\)
\(\displaystyle \displaystyle \sin^2 u + \cos^2 u = 1\)
\(\displaystyle \displaystyle 1+ \cot^2 u = \csc^2 u\)
\(\displaystyle \displaystyle \tan^2 u +1 = \sec^2 u\)
None of the above

4.

Evaluate the indefinite integral.

\(\displaystyle \int 216 \cos^4(18 x) dx \ =\)

5.

Evaluate the indefinite integral.

\(\displaystyle\int \tan^{11}\mathopen{}\left(x\right)\sec^{4}\mathopen{}\left(x\right) \, dx\) =

6.

Evaluate the integral:

\(\displaystyle \int {\frac{1 \tan^3(x)}{\cos^4(x)}}\, dx\ =\)

7.

Evaluate the integral \(\displaystyle \int {2 \cot^5(x) \sin^4(x)}\, dx\ =\)

8.

Evaluate the indefinite integral.

\(\displaystyle\int \sin(3 x)\cos(5 x)\ dx=\) \(+C\)

9.

Evaluate the indefinite integral.

\begin{equation*} \int \sec^3 x \,dx \end{equation*}

Answer: \(\ + \ C\)

10.

Match each of the trigonometric expressions below with the equivalent non-trigonometric function from the following list.

\(\cos(\theta)\) where \(x = 6 \tan \theta\)
\(\tan(\theta)\) where \(x = 6 \sin \theta\)
\(\sin(\theta)\) where \(x = 6 \tan \theta\)
\(\cos(\theta)\) where \(x = 6 \sin \theta\)

\(\displaystyle \displaystyle \frac{x}{\sqrt{36 - x^2}}\)
\(\displaystyle \displaystyle \frac{6}{\sqrt{36 + x^2}}\)
\(\displaystyle \displaystyle \frac{x}{\sqrt{36 + x^2}}\)
\(\displaystyle \displaystyle \frac{\sqrt{36 - x^2}}{6}\)
None of the above

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