(AI2) Trigonometric Integrals (Even Powers of Sine and Cosine)#

In this lesson we are going to see how to calculate integrals of the follwing form where both exponents \(m\) and \(n\) are even:

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

Lecture Videos#

Even Powers#

If both of the \(\sin x\) and \(\cos x\) terms have an even power then we:

  1. Break up our function into multiples of \(\sin^2 x\) and \(\cos^2 x\)

  2. Apply the half-angle identities.

  3. Multiply all of the terms out.

  4. Repeat the half-angle identities as needed.

Half-Angle Identities

\[\begin{split}\sin^2 \theta &=\tfrac{1}{2}(1-\cos 2\theta)\\ \cos^2 \theta &=\tfrac{1}{2}(1+\cos 2\theta)\end{split}\]

It may also be helpful to occasionally use:

Yet Another Trig Identity

\[\sin x \cos x =\tfrac{1}{2}\sin 2x\]

K-form Integrals#

We’ve mentioned these before, but it will be helpful to recall our K-form Integrals:

K-form Integrals

\[\begin{split}&\int \sin(kx)\, dx = -\tfrac{1}{k}\cos(kx)+C\\ &\int \cos(kx)\, dx = \tfrac{1}{k}\sin(kx)+C\end{split}\]

Example 4#

Integrate the following:

\[ \int_0^{\pi/2} \sin^2 x \, dx \]
Solution (Click to see the steps.)

Half-Angle: Our function is already written as a multiple of \(\sin^2 x\), so we can start by applying our half-angle identity.

\[\int_0^{\pi/2} \sin^2 x \, dx= \int_0^{\pi/2} \left(\tfrac{1}{2}-\tfrac{1}{2}\cos 2x \right)\, dx\]

Half-Angle Identity

\[\sin^2 \theta =\tfrac{1}{2}(1-\cos 2\theta)=\tfrac{1}{2}-\tfrac{1}{2}\cos 2\theta\]

Antiderivative: We’ve reduced our even power of sine, down to one of our K-form integrals. So at this point we can find the antiderivative.

\[\int_0^{\pi/2} \left(\tfrac{1}{2}-\tfrac{1}{2}\cos 2x \right)\, dx= \tfrac{1}{2}x-\tfrac{1}{2}\cdot \left(\tfrac{1}{2}\sin 2x\right) \biggr|_{x=0}^{x=\pi/2}\]

K-form Integral

\[\int \cos(kx)\, dx = \tfrac{1}{k}\sin(kx)+C\]

Evaluate: Finally we can evaluate:

\[\tfrac{1}{2}x-\tfrac{1}{4}\sin 2x \biggr|_{x=0}^{x=\pi/2} = \biggr( \tfrac{1}{2}\cdot \tfrac{\pi}{2}-\tfrac{1}{4}\sin \pi \biggr) -\biggr( \tfrac{1}{2}\cdot 0-\tfrac{1}{4}\sin 0 \biggr)\]

And if we remember that \(\sin \pi=0\) and \(\sin 0 = 0\) we can simplify this down to get:

\[\int_0^{\pi/2} \sin^2 x \, dx =\dfrac{\pi}{4}\]

Example 5#

Integrate the following:

\[ \int \sin^2 x \cos^2 x \, dx \]
Solution (Click to see the steps.)

Half-Angle: Our function is already written as a multiple of \(\sin^2 x\) and \(\cos^2 x\) terms, so we can start by applying our half-angle identities.

\[\begin{split}\int \sin^2 x \cos^2 x \, dx&= \int \tfrac{1}{2}(1-\cos 2x)\cdot \tfrac{1}{2}(1+\cos 2x)\, dx \\ &= \tfrac{1}{4}\int (1-\cos 2x)\cdot (1+\cos 2x)\, dx \end{split}\]

Half-Angle Identities

\[\begin{split}\sin^2 \theta &=\tfrac{1}{2}(1-\cos 2\theta)\\ \cos^2 \theta &=\tfrac{1}{2}(1+\cos 2\theta)\end{split}\]

Multiply: In order to continue the integration, we need to multiply out the terms in our integral:

\[\tfrac{1}{4}\int (1-\cos 2x)\cdot (1+\cos 2x)\, dx = \tfrac{1}{4}\int (1-\cos^2 2x)\, dx\]

It may help to remember FOIL when you do this:

\[(1-\cos 2x)\cdot (1+\cos 2x) = 1+\cos 2x -\cos 2x- \cos^2 2x\]

Half-Angle: Doing the multiplication in the last step increased the exponent on our cosine term (back up to a square). So we need to use our half-angle identities again.

\[\begin{split}\tfrac{1}{4}\int (1-\cos^2 2x)\, dx &= \tfrac{1}{4}\int 1-\left(\tfrac{1}{2}+\tfrac{1}{2}\cos 2(2x)\right)\, dx\\ &=\tfrac{1}{4}\int \left( \tfrac{1}{2}-\tfrac{1}{2}\cos 4x \right)\, dx\\ &=\tfrac{1}{8}\int \left( 1-\cos 4x \right)\, dx\end{split}\]

Half-Angle Identity

\[\cos^2 \theta =\tfrac{1}{2}(1+\cos 2\theta)=\tfrac{1}{2}+\tfrac{1}{2}\cos 2\theta\]

Antiderivative: It took us a few tries, but at this point we’ve reduced our even powers of sine and cosine, down to one of our K-form integrals. So now we’re ready to integrate:

\[\tfrac{1}{8}\int \left( 1-\cos 4x \right)\, dx = \tfrac{1}{8}\left(x-\tfrac{1}{4}\sin 4x\right) + C\]

K-form Integral

\[\int \cos(kx)\, dx = \tfrac{1}{k}\sin(kx)+C\]

Example 6#

Integrate the following:

\[ \int \sin^4 x \, dx \]
Solution (Click to see the steps.)

Break Up: The first thing we need to do here is rewrite our function as multiples of \(\sin^2 x\) and \(\cos^2 x\) terms.

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

Half-Angle: Next we can apply our half-angle identities.

\[\begin{split}\int \sin^2 x \sin^2 x \, dx&= \int \tfrac{1}{2}(1-\cos 2x)\cdot \tfrac{1}{2}(1-\cos 2x)\, dx \\ &= \tfrac{1}{4}\int (1-\cos 2x)\cdot (1-\cos 2x)\, dx \end{split}\]

Half-Angle Identities

\[\begin{split}\sin^2 \theta &=\tfrac{1}{2}(1-\cos 2\theta)\\ \cos^2 \theta &=\tfrac{1}{2}(1+\cos 2\theta)\end{split}\]

Multiply: In order to continue the integration, we need to multiply out the terms in our integral:

\[\begin{split}&= \tfrac{1}{4}\int (1-\cos 2x)\cdot (1-\cos 2x)\, dx\\ & = \tfrac{1}{4}\int \bigg(1-2\cos 2x+\cos^2 2x\bigg)\, dx\end{split}\]

It may help to remember FOIL when you do this:

\[(1-\cos 2x)\cdot (1-\cos 2x) = 1-\cos 2x -\cos 2x+\cos^2 2x\]

Half-Angle: Doing the multiplication in the last step increased the exponent on our cosine term (back up to a square). So we need to use our half-angle identities again.

\[\begin{split}& = \tfrac{1}{4}\int \bigg(1-2\cos 2x+\cos^2 2x\bigg)\, dx\\ & = \tfrac{1}{4}\int \bigg(1-2\cos 2x+\tfrac{1}{2}+\tfrac{1}{2}\cos 4x\bigg)\, dx\\ & = \tfrac{1}{4}\int \bigg(\tfrac{3}{2}-2\cos 2x+\tfrac{1}{2}\cos 4x\bigg)\, dx \end{split}\]

Half-Angle Identity

\[\cos^2 2x =\tfrac{1}{2}(1+\cos 2(2x))=\tfrac{1}{2}+\tfrac{1}{2}\cos 4x\]

Antiderivative: It took us a few tries, but at this point we’ve reduced our even powers of sine and cosine, down to one of our K-form integrals. So now we’re ready to integrate:

\[\begin{split}& = \tfrac{1}{4}\int \bigg(\tfrac{3}{2}-2\cos 2x+\tfrac{1}{2}\cos 4x\bigg)\, dx\\ &= \tfrac{1}{4}\bigg(\tfrac{3}{2}x-\sin 2x+\tfrac{1}{8}\sin 4x\bigg) + C\end{split}\]

And as always when we do an indefinite integral, we need to remember the +C at the end.

K-form Integral

\[\int \cos(kx)\, dx = \tfrac{1}{k}\sin(kx)+C\]

Different Angles#

Sometimes we will come across trig integrals where the inner functions, the angles, are different. In these cases it will be helpful to use:

Different Angle Trig Identities

\[\begin{split}\sin A \cos B &=\tfrac{1}{2}\bigg(\sin(A-B)+\sin(A+B)\bigg)\\ \sin A \sin B &=\tfrac{1}{2}\bigg(\cos(A-B)-\cos(A+B)\bigg)\\ \cos A \cos B &=\tfrac{1}{2}\bigg(\cos(A-B)+\cos(A+B)\bigg)\end{split}\]

Example 7#

Integrate the following:

\[ \int \sin3x \cos7x \, dx \]
Solution (Click to see the steps.)

Different Angle: The trig functions have different angles so we can apply our different angle identity.

\[\begin{split}\int \sin3x \cos7x \, dx &= \tfrac{1}{2}\int \bigg(\sin(3x-7x)+\sin(3x+7x)\bigg)\, dx \\ &= \tfrac{1}{2}\int \bigg(\sin(-4x)+\sin(10x)\bigg)\, dx \end{split}\]

Different Angle Trig Identities

\[\begin{split}\sin A \cos B =\tfrac{1}{2}\bigg(\sin(A-B)+\sin(A+B)\bigg)\\ \end{split}\]

K-form Integrals: Each of these is now one of our k-form integrals, which we can just integrate:

\[\begin{split}&= \tfrac{1}{2}\int \bigg(\sin(-4x)+\sin(10x)\bigg)\, dx \\ &= \tfrac{1}{2}\bigg(\tfrac{1}{4}\cos(-4x)-\tfrac{1}{10}\cos(10x)\bigg)+C \end{split}\]

And as always when we do an indefinite integral, we need to remember the +C at the end.

K-form Integrals

\[\int \sin(kx)\, dx = -\tfrac{1}{k}\cos(kx)+C\]