(AI1) Trigonometric Integrals (Odd Powers of Sine and Cosine)

In this lesson we are going to see how to calculate integrals of the follwing form where at least one of the exponents \(m\) or \(n\) (or both) is odd:

\[ \int \sin^m x \cos^n x \, dx \]

Lecture Videos

Odd Powers

If one (or both) of the \(\sin x\) or \(\cos x\) terms has an odd power then we:

1. Factor off one trig term from this odd power and put it next to the \(dx\).

2. Convert all remaining terms to the opposite trig function.

3. Set \(u\) equal to the opposite trig function.

4. Finish integrating using Substitution Rule.

Trigonometric Identity

\[\sin^2 \theta +\cos^2 \theta =1\]

Example 1

Integrate the following:

\[ \int \cos^3 x \, dx \]

Solution (Click to see the steps.)

Step 1

Separate: We see that cosine has the odd power, so we begin by separating off one of these cosine factors:

\[\int \cos^3 x \, dx = \int \cos^2 x \cdot \cos x\, dx\]

Step 2

Convert: Next, we convert all remaining terms to the opposite trig function. In this case we convert all remaining cosine terms into sine terms.

\[\int \cos^2 x \cdot \cos x\, dx = \int \left(1-\sin^2 x\right) \cdot \cos x\, dx \]

This is where we use the trig identity: \(\sin^2 \theta +\cos^2 \theta =1\)

Step 3

Choose u: Now we’re ready for a \(u\)-substitution, so we choose \(u\) to be the opposite trig function. In this case we choose:

\[ u=\sin x\]

and then we calculate \(du\) to get:

\[ du = \cos x \, dx\]

And then substituting our \(u\) and \(du\) into the integral gives us:

\[\int \left(1-\sin^2 x\right) \cdot \cos x\, dx = \int \left(1-u^2 \right) \, du\]

Step 4

Finish It! And finally we finish the integration:

\[\begin{split}\int \left(1-u^2 \right) \, du &= u-\tfrac{1}{3}u^3+C \\ &= \sin x-\tfrac{1}{3}(\sin x)^3+C \\\end{split}\]

(Don’t forget \(+C\) for indefinite integrals!)

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Example 2

Integrate the following:

\[ \int \sin x \cos x \, dx \]

Solution (Click to see the steps.)

Step 1

Separate: We see that cosine has an odd power, so we begin by separating off one of these cosine factors:

\[\int \sin x \cos x \, dx= \int \sin x \cdot \cos x\, dx\]

Step 2

Convert: Next, we would need to convert any remaining cosine terms into sine terms (besides that one copy of \(\cos x\) we moved next to the \(dx\)). But our integral is already in this form!

\[\int \sin x \cdot \cos x\, dx\]

So we’re all set with this step.

Step 3

Choose u: Now we’re ready for a \(u\)-substitution, so we choose \(u\) to be the opposite trig function. In this case we choose:

\[ u=\sin x\]

and then we calculate \(du\) to get:

\[ du = \cos x \, dx\]

And then substituting our \(u\) and \(du\) into the integral gives us:

\[\int \sin x \cdot \cos x\, dx = \int u \, du\]

Step 4

Finish It! And finally we finish the integration:

\[\begin{split}\int u \, du &= \tfrac{1}{2}u^2+C \\ &= \tfrac{1}{2}(\sin x)^2+C \\\end{split}\]

(Don’t forget \(+C\) for indefinite integrals!)

Example 3

Integrate the following:

\[ \int \sin^5 x \cos^2 x \, dx \]

Solution (Click to see the steps.)

Step 1

Separate: We see that sine has an odd power, so we begin by separating off one of these sine factors:

\[\int \sin^5 x \cos^2 x \, dx= \int \sin^4 x \cdot \cos^2 x\cdot \sin x \, dx\]

and we move it next to the \(dx\).

Step 2

Convert: Next, we need to convert those remaining sine terms into cosine terms (besides that one copy of \(\sin x\) we moved next to the \(dx\)).

\[\int \sin^4 x \cdot \cos^2 x\cdot \sin x \, dx\]

This is where we use the trig identity: \(\sin^2 \theta +\cos^2 \theta =1\)

\[\begin{split}& = \int \sin^2 x\cdot \sin^2 x \cdot \cos^2 x\cdot \sin x \, dx \\ &= \int \left(1-\cos^2 x\right)\cdot \left(1-\cos^2 x\right) \cdot \cos^2 x\cdot \sin x \, dx \end{split}\]

Step 3

Choose u: Now we’re ready for a \(u\)-substitution, so we choose \(u\) to be the opposite trig function. In this case we choose:

\[ u=\cos x\]

and then we calculate \(du\) to get:

\[ du = -\sin x \, dx \longrightarrow (-1)du= \sin x \, dx\]

And then substituting our \(u\) and \((-1)du\) into the integral gives us:

\[\begin{split}&= \int \left(1-\cos^2 x\right)\cdot \left(1-\cos^2 x\right) \cdot \cos^2 x\cdot \sin x \, dx \\ &= \int \left(1-u^2\right)\cdot \left(1-u^2\right) \cdot u^2 \cdot (-1) du\end{split}\]

Step 4

Finish It! In order to finish the integration we first need to multiply everything out:

\[\begin{split}&= \int \left(1-u^2\right)\cdot \left(1-u^2\right) \cdot u^2 \, (-1) du\\ &= \int \left(1-2u^2+u^4\right) \cdot u^2 \, (-1) du\\ &= \int \left(u^2-2u^4+u^6\right) \, (-1) du\\ &= \int \left(-u^2+2u^4-u^6\right) \, du\end{split}\]

And finally we can find the antiderivative:

\[\begin{split}&= -\tfrac{1}{3}u^3+\tfrac{2}{5}u^5-\tfrac{1}{7}u^7 +C\\ &= -\tfrac{1}{3}\cos^3 x+\tfrac{2}{5}\cos^5 x-\tfrac{1}{7}\cos^7 x +C \end{split}\]

(Don’t forget \(+C\) for indefinite integrals!)