Part 7: Powers of Trigonometric Functions | Beginners Guide to Year 12 Maths Ext 2

In this article, we will go through integrating powers of sine, cosine, sec, cosec, tan and cot! We also provide you with a FREE worksheet to help you practice your skills!

Are you struggling with powers of trigonometric functions? Well, you came to the right place! In this article, we will go through everything you need to know to ace powers of trigonometric functions.

 

 

In this article, we’ll discuss:

 

 

NESA Syllabus Outcomes:

  • MEX-C1 Students find and evaluate indefinite and definite integrals using the method of integration by substitution, where the substitution may or may not be given. They evaluate integrals using the method of integration by parts.

 

 

Assumed Knowledge:

Students should be familiar with how to integrate and differentiate a range of trigonometric functions.

Students should also know how to do integration by parts. Students should also be familiar with trigonometric identities.

 

Reverse Chain Rule

Recall the reverse chain rule where:

\( \int \big{(} f(x) \big{)}^{n} \times f'(x) dx \ = \ \frac{1}{n+1} \big{(} f(x) \big{)} ^{n+1} + C \)

 

Using this rule, we can derive a variety of trigonometric integrals:

\begin{align*}
&\int sin^{n}(x) \ cos(x) \ dx \ = \ \frac{sin^{n+1}(x)}{n+1} + C \\
&\int cos^{n}(x) \ sin(x) \ dx \ = \ – \frac{cos^{n+1}(x)}{n+1} + C \\
&\int tan^{n}(x) \ sec^{2} (x) \ dx \ = \ \frac{tan^{n+1}(x)}{n+1} + C \\
&\int cot^{n}(x) \ cosec^{2}(x) \ dx \ = \ – \frac{cot^{n+1}(x)}{n +1} + C \\
&\int sec^{n}(x) \ tan(x) \ dx \ = \ \frac{sec^{n}(x)}{n} + C \\
&\int cosec^{n}(x) \ cot(x) \ dx \ = \ – \frac{cosec^{n}(x)}{n} + C
\end{align*}

Many questions that involve powers of trigonometric function will be focused on attaining an integral in the form of the ones above.

 

 

Integrating \( sin^{n}(x) \) and \( cos^{n}(x) \) :

\begin{align*}\int sin^{n}(x) dx
\end{align*}

If \( n \) is an odd integer, factor out \( sin(x) \) then use the identity \( \int sin^{2}(x) \ = \ 1 – cos^{2}(x) \) for the term left after one power has been factored out.

The integral you get will be in the form of the ones in the reverse chain rule section.

 

Example 1

\( \int sin^{5} (x) \ dx \)

\begin{align*}
\int sin^{5} (x) \ dx \ &= \ \int \big{(} 1-cos^{2}(x) \big{)} ^{2} \ sin(x) \ dx \\
&= \int \big{(} 1 \ – \ 2 \ cos^{2} (x) + cos^{4} (x) \big{)} \ sin(x) \ dx \\
&= \ – cos(x) + \frac{2}{3} cos^{3} (x) \ – \frac{1}{5} cos^{5} (x) + C
\end{align*}

 

If \( n \) is an even integer, use \( sin^{2} (x) \ = \ \frac{1}{2} \big{(} 1 – cos(2x) \big{)} \) to turn all of the sines into cosines of a double angle.

Once you have done this, expand the bracket and integrate. You may need to apply more rules before you can integrate the expression you attain.

\begin{align*}
\int cos^{n}(x) \ dx
\end{align*}

 

If \( n \) is an odd integer, factor out then use the identity \( cos^{2}(x) \ = \ 1 – sin^{2}(x) \) for the term left after one power has been factored out.

The integral you get will be in the form of the ones in the reverse chain rule section.

 

If \( n \) is an even integer, use \( cos^{2}(x) \ = \ \frac{1}{2} \big{(} 1 + cos(2x) \big{)} \) to turn all of the cosines into sines of a double angle.

Once you have done this, expand the bracket and integrate. You may need to apply more rules before you can integrate the expression you attain.

 

Example 2 :

\( \int cos^{4} (x) \ dx \)

\begin{align*}
\int cos^{4} \ dx \ &= \ \int \bigg{(} \frac{1}{2} \big{(} 1 + cos(2x) \big{)} \bigg{)} ^{2} \ dx \\
&= \frac{1}{4} \int \big{(} 1 + 2 cos(2x) \ + \ cos^{2} (2x) \big{)} \ dx
\end{align*}

We notice here that we need to apply the rule again on \( cos^{2} (2x) \)

\begin{align*}
&= \frac{1}{4} \int \big{(} 1 + 2cos(2x) + \frac{1}{2} ( 1 + cos(4x) \big{)} \ dx \\
&= \frac{1}{8} \int \big{(} 3 + 4cos(2x) \ + \ cos(4x) \big{)} \ dx \\
&= \frac{3}{8}x + \frac{1}{4} sin(2x) + \frac{1}{32} sin(4x) \ + \ C
\end{align*}

 

Note that integrating powers of \( sin(x) \) and \( cos(x) \)can also be done through integration by parts!

 

 

Want to put your Powers of Trig Function skills to practice?

 

 

Integrating \( sec^{n}(x) \) and \( cosec^{n}(x) \)

\begin{align*}
\int sec^{n} (x) \ dx
\end{align*}

If \( n \) is an even integer, factor out \( sex^{2} (x) \) then use the identity \( sex^{2} (x) = 1 + tan^{2}(x) \) for the term left after two powers have been factored out.

Expand this integral.

The integral you get will be in the form of the ones in the reverse chain rule section.

 

Example

\( \int sec^{6}(x) \ dx \)

\begin{align*}
\int sec^{6}(x) \ dx \ &= \ \int \big{(} 1 + tan^{2} (x) \big{)} ^{2} sec^{2} (x) \ dx \\
&= \int \big{(} 1 + tan^{2}(x) + tan^{4}(x) \big{)} sec^{2} \ dx \\
&= tan(x) + \frac{2}{3} tan^{3} (x) + \frac{1}{5} tan^{5} + C
\end{align*}

 

If \( n \) is an odd integer, use integration by parts. See the reduction formula blog post for more information.

Note that \( \int sec(x) \ dx \ = \ ln \ | sec(x) + tan(x) | + C  \)

\begin{align*}
\int cosec^{n} (x) \ dx
\end{align*}

Use complementary angles, and a \( u \) substitution of \( u = \frac{\pi}{2} – x \) to change all \( cosec(x) \) into \( sec(x) \).

Once you have obtained an integral in terms of powers of \( sec(x) \) use the aforementioned process to integrate!

 

 

Integrating \( tan^{n}(x) \) and \( cot^{n}(x) \)

\begin{align*}
\int tan^{n} (x) dx
\end{align*}

Note that \( \int tan(x) dx \ = \ -ln | cos(x) | + C \).

If \( n \) is an even OR odd integer, factor out a \( tan^{2} (x) \) and write it as \( sec^{2}(x) – 1 \).

Expand out the integral.

You will obtain an integral in the form of \( tan^{n}(x) \ sec^{2} (x) \) which you can apply the reverse chain rule on, as well as a lower power of to \( tan(x) \) integrate (repeat this process on this integral).

 

Example:

\( \int tan^{5} (x) \ dx \)

\begin{align*}
\int tan^{5} (x) \ dx &= \int tan^{3} (x) \big{(} sec^{2} (x) – 1 \big{)} \ dx \\
&= \frac{1}{4} tan^{4} (x) \ – \int tan^{3} (x) \ dx \\
&= \frac{1}{4} tan^{4}(x) \ – \int tan(x) \big{(} sec^{2} (x) – 1 \big{)} \ dx \\
&= \frac{1}{4} tan^{4}(x) \ – \frac{1}{2} tan^{2} (x) – ln \ |cos(x)| + C
\end{align*}

 

Use complementary angles, and a \( u = \frac{\pi}{2} – x \) substitution of to change all \( cot(x) \) into\( tan(x) \).

Once you have obtained an integral in terms of powers of use the aforementioned process to integrate!

 

Mixed expressions

\begin{align*}
\int sin^{n}(x) \ cos^{m} (x) \ dx
\end{align*}

To approach an integral in this form, apply one of the following strategies that corresponds to the appropriate case.

 

1. If \( m \) is odd and \( n \) is even:

Factor out a \( cos(x) \), then convert the remaining \( cos(x) \) terms to \( sin(x) \) using \( cos^{2}(x) = 1 – sin^{2}(x)\).

Expand this integral out and the integral you attain can be integrated using the reverse chain rule.

 

2. If \( m \) is even and \( n \) is odd:

Factor out a \( sin(x) \), then convert the remaining \( sin(x) \) terms to \( cos(x) \) using \( sin(x)^{2}(x) \ = \ 1 – cos^{2}(x) \).

Expand this integral out and the integral you attain can be integrated using the reverse chain rule.

 

3. If \( m \) and \( n \) are odd:

You can use either of the two previously mentioned methods to integrate.

 

4. If \( m \) and \( n \) are even:

Express \( sin(x) \) as \( \frac{1}{2} \big{(} 1 – cos(2x) \big{)} \) and as in your integral.

This will change your integral into a sum of integrals in the form \( \int cos^{k} (2x) \ dx \) which you can integrate using the methods discussed previously.

 

Example 1

\( \int sin^{3} (x) \ cos^{4} (x) \ dx \)

 

Solution 1

\begin{align*}
\int sin^{3} \ cos^{4} \ dx &= \int sin(x) \big{(} 1 – cos^{2} (x) \big{)} cos^{4}(x) \ dx \\
&= \int \big{(} sin(x) \ cos^{4} (x) – sin{x} \ cos^{6}(x) \big{)} \\
&= \ – \frac{cos^{5} (x) }{5} + \frac{cos^{7}(x)}{7} + C
\end{align*}

 

Example 2

\( \int sin^{2} (x) \ cos^{2}(x) \ dx \)

 

Solution 2

\begin{align*}
\int sin^{2}(x) cos^{2}(x) \ dx &= \int \frac{1}{2} \big{(} 1 – cos(2x) \big{)} \cdot \frac{1}{2} \big{(} 1 + cos(2x) \big{)} \ dx \\
&= \int \frac{1}{4} \big{(} 1 – cos^{2} (2x) \big{)} \ dx \\
&= \int \frac{1}{4} \bigg{(} 1 – \frac{1}{2} \big{(} 1 + cos(4x) \big{)} \bigg{)} \ dx \\
&= \int \bigg{(} \frac{1}{8} – \frac{1}{8} cos(4x) \bigg{)} \ dx \\
&= \frac{1}{8} x \ – \frac{1}{32} sin(4x) \ + \ C
\end{align*}

 

Concept check questions

1. \( \int cos^{3} (2x) \ dx \)

 

2. \( \int sec^{3} (x) \ dx \)

 

3. \( \int tan^{4} (x) \ dx \)

 

4. \( \int sin^{4} (x) \ cos^{2} (x) \ dx \)

 

 

Concept check solutions

1. \( \frac{sin(2x)}{2} – \frac{sin^{3}(2x)}{6} + C \)

 

2. \( \frac{1}{2} sec(x) \ tan(x) + \frac{1}{2} ln \ | sec(x) \ + \ tan(x) | \ + C \)

 

3. \( \frac{1}{3} tan^{3} (x) – tan(x) + x + C \)

 

4. \( \frac{x}{16} – \frac{sin(4x)}{64} – \frac{sin^{3}(2x)}{48} + C \)

 

 

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