In engineering applications you sometimes encounter integrals of the form
\[\int \cos\;(\alpha t + \phi_1)~\cos\;(\beta t + \phi_2)~\dt \nonumber \]
where \(\alpha t + \phi_1\) and \(\beta t + \phi_2\) are different angles (e.g. when the voltage and current are out of phase in an AC circuit). In general, integrals involving products of sines and cosines with “mixed” angles can be simplified with the useful product-to-sum formulas:2
Example \(\PageIndex{1}\): trigint1
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Solution
Evaluate \(~\displaystyle\int 0.5\,\sin\,x\;\sin\,12x~\dx~\).
Solution: Using the product-to-sum formula ([eqn:p2ssinsin]) with \(A=x\) and \(B=12x\),
\[\begin{aligned} \sin\;A~\sin\;B ~&=~ -\tfrac{1}{2}\;(\cos\;(A+B) ~-~ \cos\;(A-B))\\ \sin\,x~\sin\,12x ~&=~ -\tfrac{1}{2}\;(\cos\;(x+12x) ~-~ \cos\;(x-12x))\\ \sin\,x~\sin\,12x ~&=~ -\tfrac{1}{2}\;(\cos\,13x ~-~ \cos\,11x)\end{aligned} \nonumber \]
since \(\cos\,(-11x) = \cos\,11x\). Then
\[\begin{aligned} \int 0.5\,\sin\,x\;\sin\,12x~\dx ~&=~ -\frac{1}{4}\,\int (\cos\,13x ~-~ \cos\,11x)~\dx\
\[6pt] &=~ -\frac{1}{52}\,\sin\,13x ~+~ \frac{1}{44}\,\sin\,11x ~+~ C\end{aligned} \nonumber \]
Notice how the product-to-sum formula turned an integral of products of sines into integrals of individual cosines, which are easily integrated. The integrand is an example of a modulated wave, commonly used in electronic communications (e.g. radio broadcasting). The graph is shown below:
The curves \(y=\pm 0.5 \sin\,x\) (shown in dashed lines) form an amplitude envelope for the modulated wave.
On occasion you might need to integrate trigonometric functions raised to powers higher than two. For the sine function raised to odd powers of the form \(2n+1\) (for \(n \ge 1\)), the trick is to replace \(\sin^2 x\) by \(1 - \cos^2 x\), so that
\[\begin{aligned} \int \sin^{2n+1} x~\dx ~&=~ \int (\sin^2 x)^n \,\sin\,x~\dx\
\[6pt] &=~ \int (1 - \cos^2 x)^n \,\sin\,x~\dx\
\[6pt] &=~ \int p(u)~\du\end{aligned} \nonumber \]
where \(p(u)\) is a polynomial in the variable \(u=\cos\,x\), and the single \(\sin\,x\) is now part of \(\du = -\sin\,x\;\dx\). You can then use the Power Formula to integrate that polynomial.
Example \(\PageIndex{1}\): trigint2
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Solution
Evaluate \(~\displaystyle\int \sin^3 x~\dx~\).
Solution: Let \(u=\cos\,x\) so that \(\du=-\sin\,x\;\dx\):
\[\begin{aligned} \int \sin^3 x~\dx ~&=~ \int (\sin^2 x) \;\sin\,x~\dx\
\[6pt] &=~ \int (1 - \cos^2 x) \;\sin\,x~\dx\
\[6pt] &=~ \int (1 - u^2)\;(-\du)\
\[6pt] &=~ \int (u^2 - 1)\;\du\
\[6pt] &=~ \frac{1}{3}\,u^3 ~-~ u ~+~ C\
\[4pt] &=~ \frac{1}{3}\,\cos^3 x ~-~ \cos\,x ~+~ C\end{aligned} \nonumber \]
In general \(\int \sin^{2n+1}\,x\;\dx\) will be a polynomial of degree \(2n+1\) in terms of \(\cos\,x\). Similarly, use \(\cos^2 x = 1 - \sin^2 x\) to integrate odd powers of \(\cos\,x\), with the substitution \(u=\sin\,x\):
\[\begin{aligned} \int \cos^{2n+1} x~\dx ~&=~ \int (\cos^2 x)^n \,\cos\,x~\dx\
\[6pt] &=~ \int \underbracket[0.3pt]{(1 - \sin^2 x)^n\vphantom{\frac{1}{2}}}_{p(u)} \;\underbracket[0.3pt]{\cos\,x~\dx\vphantom{\frac{1}{2}}}_{\du}\end{aligned} \nonumber \]
Integrals of the form \(\int \sin^m\,x\;\cos^n\,x\;\dx\), where either \(m\) or \(n\) is odd, can be evaluated using the above trick for the function having the odd power.
Example \(\PageIndex{1}\): trigint3
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Solution
Evaluate \(~\displaystyle\int \sin^2 x\;\cos^3 x~\dx~\).
Solution: Replace \(\cos^2 x\) by \(1 - \sin^2 x\), then let \(u=\sin\,x\) so that \(\du=\cos\,x\;\dx\):
\[\begin{aligned} \int \sin^2 x\;\cos^3 x~~\dx ~&=~ \int \sin^2 x ~(\cos^2 x) \;\cos\,x~\dx\
\[6pt] &=~ \int \underbracket[0.3pt]{\sin^2 x ~(1 - \sin^2 x)\vphantom{\frac{1}{2}}}_{p(u)} \;\underbracket[0.3pt]{\cos\,x~\dx\vphantom{\frac{1}{2}}}_{\du}\
\[3pt] &=~ \int (u^2 - u^4)\;\du\
\[6pt] &=~ \frac{1}{3}\,u^3 ~-~ \frac{1}{5}\,u^5 ~+~ C\
\[4pt] &=~ \frac{1}{3}\,\sin^3 x ~-~ \frac{1}{5}\,\sin^5 x ~+~ C\end{aligned} \nonumber \]
For even powers of \(\sin\,x\) or \(\cos\,x\). You would replace \(\sin^2 x\) or \(\cos^2 x\) with either
\[\sin^2 x ~=~ \frac{1 \;-\; \cos\,2x}{2} \quad\quad\text{or}\quad\quad \cos^2 x ~=~ \frac{1 \;+\; \cos\,2x}{2} ~, \nonumber \]
respectively, as often as necessary, then proceed as before if odd powers occur.
Example \(\PageIndex{1}\): trigint4
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Solution
Evaluate \(~\displaystyle\int \sin^4 x ~\dx~\).
Solution: Replace \(\sin^2 x\) by \(\frac{1 - \cos\,2x}{2}\):
\[\begin{aligned} \int \sin^4 x~\dx ~&=~ \int \left(\frac{1 \;-\; \cos\,2x}{2}\right)^2 \,\dx\
\[6pt] &=~ \frac{1}{4}\,\int (1 ~-~ 2\,\cos\,2x ~+~ cos^2 2x)~\dx\
\[6pt] &=~ \frac{1}{4}\,\int \left(1 ~-~ 2\,\cos\,2x ~+~ \frac{1 \;+\; \cos\,4x}{2}\right)~\dx\
\[6pt] &=~ \frac{1}{8}\,\int (3 ~-~ 4\,\cos\,2x ~+~ \cos\,4x)~\dx\
\[6pt] &=~ \frac{3x}{8} ~-~ \frac{1}{4}\,\sin\,2x ~+~ \frac{1}{32}\,\sin\,4x ~+~ C\end{aligned} \nonumber \]
Similar methods can be used for integrals of the form \(\;\int \sec^m\,x\;\tan^n\,x\;\dx\;\) when either \(m\) is even or \(n\) is odd. For an even power \(m = 2k+2\), use \(\sec^2 x = 1 + \tan^2 x\) for all but two of the \(m\) powers of \(\sec\,x\), then use the substitution \(u=\tan\,x\), so that \(\du=\sec^2 x\;\dx\). This results in an integral of a polynomial \(p(u)\) in terms of \(u=\tan\,x\):
\[\begin{aligned} \int \sec^{2k+2} x ~\tan^n x~\dx ~&=~ \int (\sec^2 x)^k \,\sec^2 x~\tan^n x~\dx\
\[6pt] &=~ \int \underbracket[0.3pt]{(1 + \tan^2 x)^k\,\tan^n x\vphantom{\frac{1}{2}}}_{p(u)} \;\underbracket[0.3pt]{\sec^2 x~\dx\vphantom{\frac{1}{2}}}_{\du}\end{aligned} \nonumber \]
Likewise for an odd power \(n=2k+1\), use \(\tan^2 x = \sec^2 x - 1\) for all but one of the \(n\) powers of \(\tan\,x\), then use the substitution \(u=\sec\,x\), so that \(\du=\sec\,x\;\tan\,x\;\dx\). This results in an integral of a polynomial \(p(u)\) in terms of \(u=\sec\,x\):
\[\begin{aligned} \int \sec^m x ~\tan^{2k+1} x~\dx ~&=~ \int \sec^{m-1} x~\sec\,x \,~(\tan^2 x)^k ~\tan\,x~\dx\
\[6pt] &=~ \int \underbracket[0.3pt]{\sec^{m-1} x~(\sec^2 x - 1)^k\vphantom{\frac{1}{2}}}_{p(u)} \;\underbracket[0.3pt]{\sec\,x~\tan\,x\dx\vphantom{\frac{1}{2}}}_{\du}\end{aligned} \nonumber \]
Mimic the above procedure for integrals of the form \(\;\int \csc^m\,x\;\cot^n\,x\;\dx\;\) when either \(m\) is even or \(n\) is odd, using the identity \(\csc^2 x = 1 + \cot^2 x\) in a similar manner.
Example \(\PageIndex{1}\): trigint5
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Solution
Evaluate \(~\displaystyle\int \sec^4 x ~\tan\,x~\dx~\).
Solution: Use \(\sec^2 x = 1 + \tan^2 x\) for one \(\sec^2 x\) term, then substitute \(u=\tan\,x\), so that \(\du=\sec^2 x\;\dx\):
\[\begin{aligned} \int \sec^4 x ~\tan\,x~\dx ~&=~ \int \sec^2 x~\sec^2 x~\tan\,x~\dx\
\[6pt] &=~ \int (1 + \tan^2 x)\,\tan\,x~\sec^2 x~\dx\
\[6pt] &=~ \int (1 + u^2)\,u~\du\
\[6pt] &=~ \int (u + u^3)~\du\
\[6pt] &=~ \frac{1}{2}\,u^2 ~+~ \frac{1}{4}\,u^4 ~+~ C\\ &=~ \frac{1}{2}\,\tan^2 x ~+~ \frac{1}{4}\,\tan^4 x ~+~ C\end{aligned} \nonumber \]
For some trigonometric integrals try putting everything in terms of sines and cosines.
Example \(\PageIndex{1}\): trigint6
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Solution
Evaluate \(~\displaystyle\int \dfrac{\cot^4 x}{\csc^5 x}\;\dx~\).
Solution: Put \(\cot\,x\) and \(\csc\,x\) in terms of \(\sin\,x\) and \(\cos\,x\):
\[\begin{aligned} \int \frac{\cot^4x}{\csc^5 x}\;\dx ~&=~ \int \frac{\cos^4x~\sin^5 x}{\sin^4 x}\;\dx\
\[6pt] &=~ \int \cos^4 x~\sin\,x~\dx\quad\text{(now let $u=\cos\,x$, $\du=-\sin\,x~\dx$)}\
\[6pt] &=~ -\int u^4~\dx ~=~ -\frac{1}{5}\,\cos^5 x ~+~ C\end{aligned} \nonumber \]
[sec6dot2]
For Exercises 1-12, evaluate the given integral.
4
\(\displaystyle\int \sin\,2x~\cos\,5x~\dx\)
\(\displaystyle\int \cos\,2x~\cos\,5x~\dx\)
\(\displaystyle\int \sin\,2x~\sin\,5x~\dx\)
\(\displaystyle\int \cos\,2\pi x~\sin\,3\pi x~\dx\)
4
\(\displaystyle\int \sin^3 x~\cos^{3/2} x~\dx\)
\(\displaystyle\int \cos^4 x~\dx\)
\(\displaystyle\int \sin^6 x~\dx\)
\(\displaystyle\int \sin\,x~\sin\,2x~\sin\,3x~\dx\)
4
\(\displaystyle\int \sec^4 x~\dx\vphantom{\dfrac{\tan^3 x}{\sec^4 x}}\)
\(\displaystyle\int \sec^2 x~\tan^3 x~\dx\vphantom{\dfrac{\tan^3 x}{\sec^4 x}}\)
\(\displaystyle\int \dfrac{\tan^3 x}{\sec^4 x}\;\dx\)
\(\displaystyle\int \dfrac{\dx}{\csc^2 x ~\cot\,x}\vphantom{\dfrac{\tan^3 x}{\sec^4 x}}\)
[[1.]]
Evaluate \(~\int \sin^3 x~\cos^3 x~\dx\) in two different ways:
- Use \(\sin^3 x = (1 - \cos^2 x)\;\sin\,x\) and the substitution \(u=\cos\,x\).
- Use \(\cos^3 x = (1 - \sin^2 x)\;\cos\,x\) and the substitution \(u=\sin\,x\).
Are the answers from parts(a) and (b) equivalent? Explain.
Evaluate \(~\int \sec^4 x ~\tan\,x~\dx~\) by using \(\sec^4 x ~\tan\,x \;=\; \sec^3 x ~(\sec\,x~\tan\,x)\) and the substitution \(u=\sec\,x\). Is your answer equivalent to the answer in Example
Example \(\PageIndex{1}\): trigint5
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Solution
? Explain.
Show that \(\displaystyle\int \dfrac{4\,\tan\,x~(1 - \tan^2 x)}{(1 +\tan^2 x)^2}\;\dx ~=~ -\frac{1}{4}\,\cos\,4x ~+~ C\).
The autocorrelation function \(R_x(\tau)\) of the periodic function \(x(t) = A\,\cos\,(\omega t + \theta)\) is given by
\[R_x(\tau) ~=~ \frac{\omega}{2\pi}\,\int_0^{2\pi/\omega} x(t)\;x(t-\tau)~\dt \nonumber \]
where \(A\), \(\omega\) and \(\theta\) are constants. Show that
\[R_x(\tau) ~=~ \frac{A^2}{2}\,\cos\,\omega\tau ~. \nonumber \]
