AC Integration with Trigonometric Functions

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 (5 x)

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\int \sin^{2} (5 x) \cos (5 x) d x =

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We see that

\int \sin^{2} (5 x) \cos^{3} (5 x) d x

is solved similarly, because

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the exponent of $\sin$ is odd.
the exponent of $\cos$ is odd.
None of the above

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So we’ll still use u-substitution with

u = \sin (5 x),

after first rewriting

\cos^{3} (5 x) = \cos^{2} (5 x) \cos (5 x)

and using a trig identity.

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Select the trigonometric identity needed to solve this problem.

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

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Finally, put it all together to calculate

\int \sin^{2} (5 x) \cos^{3} (5 x) d x =

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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 (5 x)

and use the identity

\sin^{2} u + \cos^{2} u = 1

to turn the remaining

\cos^{2} (5 x)

into an expression involving

\sin^{2} (5 x) .

Thus the integral can be rewritten as

\int \sin^{2} (5 x) \cos^{3} (5 x) d x = \int \sin^{2} (5 x) \cos^{2} (5 x) \cos (5 x) d x

= \int \sin^{2} (5 x) (1 - \sin^{2} (5 x)) \cos (5 x) d x

and we use u-substitution.

Letting

u = \sin (5 x)

d u = 5 \cos (5 x) d x

and thus

d x = \frac{d u}{5 \cos (5 x)} .

This turns the integral into

\int u^{2} (1 - u^{2}) \cos (5 x) \frac{d u}{5 \cos (5 x)} = \frac{1}{5} \int u^{2} (1 - u^{2}) d u

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

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Question 9.4.2.

Consider the integral

\int - 7 \sin^{2} (6 x) d x .

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In order to integrate, the key feature is that

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

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Then the trigonometric identity needed to solve this problem is

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

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And we use this identity to rewrite the integral as

\int - 7 \sin^{2} (6 x) d x = \int

d x

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

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Question 9.4.3.

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

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

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

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Question 9.4.4.

Consider the integral

\int 5 \tan^{3} (10 x) \sec^{3} (10 x) d x .

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In order to integrate, the key feature is that

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

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Then the trigonometric identity needed to solve this problem is

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

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So to integrate, we would use u-substitution with

u = \sec (10 x) .

But first we’d have to rewrite the integral in order to have a copy of

u

’s derivative

\sec (10 x) \tan (10 x),

and use a trigonometric identity to turn the remaining even powers of

\tan (10 x)

into an expression of

\sec^{2} (10 x) .

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Question 9.4.5.

Consider the integral

\int 9 \tan^{4} (- 12 x) \sec^{4} (- 12 x) d x .

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In order to integrate, the key feature is that

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

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Then the trigonometric identity needed to solve this problem is

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

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So to integrate, we would use u-substitution with

u = \tan (- 12 x) .

But first we’d have to rewrite the integral in order to have a copy of

u

’s derivative

\sec^{2} (- 12 x),

and use a trigonometric identity to turn the remaining even powers of

\sec (- 12 x)

into an expression involving

\tan^{2} (- 12 x) .

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Exercises 9.4.2 Exercises

1.

Evaluate the indefinite integral.

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\int 3 \tan^{3} (- 13 x) \sec^{3} (- 13 x) d x =

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Select the trigonometric identity you used in solving this problem.

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

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2.

Evaluate the indefinite integral.

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\int 7 \tan^{4} (15 x) \sec^{4} (15 x) d x =

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Select the trigonometric identity you used in solving this problem.

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

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3.

Evaluate the indefinite integral.

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\int 12 \cot^{3} (- 2 x) \csc^{3} (- 2 x) d x =

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Select the trigonometric identity you used in solving this problem.

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

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4.

Evaluate the indefinite integral.

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\int 20 \cos^{4} (5 x) d x =

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5.

Evaluate the indefinite integral.

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\int \tan^{5} (x) \sec^{4} (x) d x

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6.

Evaluate the integral:

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\int \frac{- 4 \tan^{3} (x)}{\cos^{4} (x)} d x =

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7.

Evaluate the integral

\int 6 \cot^{5} (x) \sin^{4} (x) d x =

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8.

Evaluate the indefinite integral.

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\int \sin (6 x) \cos (12 x) d x =

+ C

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9.

Evaluate the indefinite integral.

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\int \sec^{3} x d x

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Answer:

+ C

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10.

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

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$\cos (θ)$ where $x = 8 \tan θ$
$\tan (θ)$ where $x = 8 \sin θ$
$\sin (θ)$ where $x = 8 \tan θ$
$\cos (θ)$ where $x = 8 \sin θ$

$\frac{x}{\sqrt{64 - x^{2}}}$
$\frac{8}{\sqrt{64 + x^{2}}}$
$\frac{x}{\sqrt{64 + x^{2}}}$
$\frac{\sqrt{64 - x^{2}}}{8}$
None of the above

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You have attempted 1 of 16 activities on this page.

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