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Calculus

Other Trigonometric Integrals

Often we have to apply trigonometric identities before integrating. This is required when integrating the form where and n are integers, \(\int \sin^m x\cos^n x\,dx\). For this form of integrals, there are three cases to consider.

Case 1. m is odd.

  1. Write \(\sin^m x\) as \(\sin^{m-1} x\sin x\)
  2. Use \(\cos^2 x+\sin^2 x=1\) to write the even power of sines in terms of cosines
  3. Use the substitution \(u=\cos x\)

Case 2. is odd.

  1. Write \(\cos^n x\) as \(\cos^{n-1} x\cos x\)
  2. Use \(\cos^2 x+\sin^2 x=1\) to write the even power of cosines in terms of sines
  3. Use the substitution \(u=\sin x\)

Case 3. Both m and are even

Use \(2\cos^2 x=1+\cos 2x\) and \(2\sin^2 x=1-\cos 2x\) to halve the even powers.

Example 1: Integrate \(\int \sin^3 x\cos^6 x\,dx\)

Solution: We can use the steps outlined in Case 1 since \(m=3\).

\[\int \sin^3 x\cos^6 x\,dx=\int \sin^2 x\sin x\cos^6 x\,dx\]

Now we substitute \(\sin^2 x=1-\cos^2 x\)

\[\int \sin^2 x\sin x\cos^6 x\,dx=\int \left(1-\cos^2 x\right)\sin x\cos^6 x\,dx=\int \sin x\cos^6 x\,dx-\int \sin x\cos^8 x\,dx\]

Now let \(u=\cos x\), then \(du=-\sin x\,dx\)

\begin{align} \int \sin x\cos^6 x\,dx-\int \sin x\cos^8 x\,dx &= \int u^6\,du-\int u^8\,du \\ &= \frac{u^7}{7} -\frac{u^9}{9}+C \\ &= \frac{\cos^7 x}{7} -\frac{\cos^9 x}{9}+C \end{align}


Example 2: Integrate \(\int \cos^2 2x\,dx\)

Solution: For this integral, we will follow the steps for Case 3 and use our double angle identities

\begin{align} \int \cos^2 2x\,dx&=\int \frac{1+\cos 4x}{2}dx\\&=\int \frac{1}{2} + \int \frac{\cos 4x}{2} \\ &=\frac{x}{2}+\frac{\sin 4x}{8}+C\end{align}

For the integrals of the form \[\int \tan^m x\sec^nx\,dx\] or \[\int \cot^m x\csc^nx\,dx\]

Case 1. and are both odd

  1. Set aside a factor \(\sec x\tan x\) or \(\csc x\cot x\)
  2. Use equations \(1+\tan^2x=\sec^2x\) to write the even power of tangents in terms of secants, or \(1+\cot^2x=\csc^2x\) to write the even power of cotangents in terms of cosecants
  3. Use the substitution \(u=\sec x\) or \(u=\csc x\)

Case 2. n is even

  1. If possible, set aside a factor \(\sec^2 x\) or \(\csc^2 x\). If not possible, use \(1+\tan^2x=\sec^2x\) to change a factor \(tan^2x\) into a term with  \(\sec^2 x\), or \(1+\cot^2x=\csc^2x\) to change a factor \(cot^2x\) into a term with \(csc^2x\). Repeat if necessary.
  2. Use \(1+\tan^2x=\sec^2x\) to write the even power of secants in terms of tangents, or \(1+\cot^2x=\csc^2x\) to write the even power of cosecants in terms of cotangents
  3. Use the substitution \(u=\tan x\) or \(u=\cot x\)

Case 3. is even, n is odd

  1. Use \(1+\tan^2x=\sec^2x\) to write the even power of tangents in terms of secants, or \(1+\cot^2x=\csc^2x\) to write the even power of cotangents in terms of cosecants
  2. Integrate the odd power of secants or cosecants by parts

Example 3: Integrate \(\int_{0}^{\pi/4} \tan x\sec^4 x\,dx\)

Solution: Let's set aside a \(\sec^2 x\) and use the identity \(\sec^2 x = 1+\tan^2 x\)

\begin{align} \int_{0}^{\pi/4} \tan x\sec^4 x\,dx&=\int_{0}^{\pi/4} \tan x\sec^2 x\left(1+\tan^2\right)dx \\ &= \int_{0}^{\pi/4} \tan x\sec^2 x +\int_{0}^{\pi/4} \tan^3 x\sec^2 x \end{align}

Let \(u=\tan x\), then \(du=\sec^2 x\,dx\)

\begin{align} \int_{0}^{\pi/4} \tan x\sec^2 x +\int_{0}^{\pi/4} \tan^3 x\sec^2 x &= \int_{x=0}^{x=\pi/4} u\,du +\int_{0}^{\pi/4} u^3 du \\ &= \left(\frac{u^2}{2} + \frac{u^4}{4}\right)_{x=0}^{x=\pi/4} \\ &=\left(\frac{\tan^2 x}{2} + \frac{\tan^4 x}{4}\right)_{0}^{\pi/4} \\&=\frac{3}{4}\end{align}

Alternative Solution: We can set aside \(\sec x\tan x\) instead and solve this another way.

\[\int_{0}^{\pi/4} \tan x\sec^4 x\,dx=\int_{0}^{\pi/4} \left(\tan x\sec x\right)\sec^3 x\,dx\]

Let \(u=\sec x\), then \(du=\sec x\tan x\,dx\). When \(x=0\), \(u=1\) and when \(x=\pi/4\), \(u=\sqrt{2}\)

\begin{align} \int_{0}^{\pi/4} \left(\tan x\sec x\right)\sec^3 x\,dx &= \int_{1}^{\sqrt{2}} u^3 du \\ &= \left(\frac{u^4}{4}\right)_{1}^{\sqrt{2}} \\&= \frac{3}{4} \end{align}


Example 4: Integrate \(\int \frac{1-\cot x}{\sin^4 x}dx\)

Solution: In some cases, you will have to use trigonometric identities and simplify before integrating

\begin{align} \int \frac{1-\cot x}{\sin^4 x}dx&=\int \frac{1}{\sin^4 x}dx- \int \frac{\cot x}{\sin^4 x}dx \\ &= \int \sec^4 x\,dx - \int \frac{\frac{\cos x}{\sin x}}{\sin^4 x}dx \\ &=  \int \left(1+\tan^2 x\right)\sec^2 x\,dx -  \int \frac{\cos x}{\sin^5 x}dx\\ &= \int\sec^2x\,dx+\int \tan^2x\sec^2 x\,dx -  \int \frac{\cos x}{\sin^5 x}dx \end{align}

For the second integral, let \(u=\tan x\), then \(du=\sec^2 x\,dx\). For the third integral, let \(v=\sin x\), then \(dv=\cos x\,dx\). 

\begin{align} \int\sec^2x\,dx+\int \tan^2x\sec^2 x\,dx -  \int \frac{\cos x}{\sin^5 x}dx &= \int\sec^2x\,dx+\int u^2\,du-\int v^{-5}\,dv \\&=\tan x+\frac{u^3}{3}-\frac{v^{-4}}{-4} +C\\&=\tan x+\frac{\tan^3x}{3}+\frac{\sin^{-4}x}{4}+C  \end{align}


Example 5: Integrate \(\int\frac{\sec^2t\tan t}{4+\sec^2t}dt\)

Solution: Sometimes you have to apply multiple techniques. In this case, the denominator is of the power -1. This suggests a logarithmic integral. So we let \(u=4+\sec^2t\), then \(du=2\sec t\left(\sec t\tan t\right)\,dt\)

\begin{align} \int\frac{\sec^2t\tan t}{4+\sec^2t}dt &= \frac{1}{2}\int\frac{du}{u} \\&= \frac{1}{2} \ln|u|+C \\&= \frac{1}{2} \ln|4+\sec^2t|+C\end{align}

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Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License.

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