Trigonometric integrals

Integrals with \(\sin\) and \(\cos\)

  • Today, we'll apply trigonometric identities and the methods of integration we've seen so far to calculate integrals involving certain combinations of trigonometric functions.
  • \(\int(\sin x)^2\cos x\,\mathrm dx\).
    • Here, we can use the substitution \(u=\sin x\): \[ \int(\sin x)^2\cos x\,\mathrm dx=\int u^2\,\mathrm du=\frac12(\sin x)^3+C. \]
  • \(\int(\sin x)^3\,\mathrm dx\).
    • We want to transform the integrand to a form where there's a polynomial of \(\cos x\) multiplied with \(\sin x\), so that we can use the Substitution rule with \(u=\cos x\).
    • To that end, we use the trigonometric identity \((\sin x)^2=1-(\cos x)^2\): \[ \int(\sin x)^3\,\mathrm dx=\int(1-(\cos x)^2)\cdot \sin x\,\mathrm dx=\int(1-(\cos x)^2)\cdot\sin\ x\,\mathrm dx=\int(-1+u^2)\,\mathrm du=-\cos x+\frac12(\cos x)^2+C. \]
  • \(\int(sin x)^2(\cos x)^3\,\mathrm dx\).
    • Here, there's an odd power of \(\cos x\), so we want to transform the integrand to a form where there's a polynomial of \(\sin x\) multiplied with \(\cos x\), so that we can use the Substitution Rule with \(u=\sin x\): \[ \int(\sin x)^2(\cos x)^3=\int(\sin x)^2(1-(\sin x)^2)\cos x\,\mathrm dx=\int u^2-u^4\,\mathrm du=\frac13(\sin x)^3-\frac15(\sin x)^5+C. \]

Integrals with \(\sin\) and \(\cos\)

  • \(\int(\cos x)^2\,\mathrm dx\).
    • To calculate this integral, we need to use a half-angle formula: \[ \int(\cos x)^2\,\mathrm dx=\frac12\int1+\cos2x\,\mathrm dx=\frac12x+\frac14\sin2x+C. \]
  • \(\int(\cos x)^4\,\mathrm dx\).
    • This integral has no odd power either, so it's best to use half-angle formulas: \[\begin{multline*} \int(\cos x)^4\,\mathrm dx=\frac14\int(1+\cos 2x)^2\,\mathrm dx=\frac14\int1+2\cos 2x+(\cos 2x)^2\,\mathrm dx\\=\frac14x+\frac14\sin2x+\frac18\int1+\cos 4x\,\mathrm dx=\frac14x+\frac14\sin2x+\frac18x+\frac{1}{32}\sin4x+C. \end{multline*}\]

Integrals with \(\sin\) and \(\cos\)

  • We're ready to state the general rules to integrate \(\int(\sin x)^m(\cos x)^n\,\mathrm dx\):
    • If \(m\) is odd, then using \((\sin x)^2=1-(\cos x)^2\) transform the integrand to have \(\sin x\) to exponent 1: \[ \int(\sin x)^m(\cos x)^n\,\mathrm dx=\int(1-(\cos x)^2)^{\frac{m-1}{2}}(\cos x)^n\sin x\,\mathrm dx \] and then use the Substitution rule with \(u=\cos x\).
    • If \(n\) is odd, then using \((\cos x)^2=1-(\sin x)^2\) transform the integrand to have \(\cos x\) to exponent 1: \[ \int(\sin x)^m(\cos x)^n\,\mathrm dx=\int(\sin x)^m(1-(\sin x)^2)^{\frac{n-1}{2}}\cos x\,\mathrm dx \] and then use the Substitution rule with \(u=\sin x\).
    • If both \(m\) and \(n\) are even, then use the half-angle formulas: \[ (\sin x)^2=\frac{1-\cos2x}{2}\,(\cos x)^2=\frac{1+\cos 2x}{2},\,\sin x\cos x=\frac{\sin 2x}{2}. \]
  • Exercises. 7.2: 2, 4, 6, 10, 14, 18.

\(\int(\tan x)^m(\sec x)^n\,\mathrm dx\)

  • If \(n\) is even.
    • You can use \((\sec x)^2=1+(\tan x)^2\) to get an expression of \(\tan x\) multiplied by \((\sec x)^2\)
    • Since \((\tan x)'=(\sec x)^2\), you can now substitute \(u=\tan x\).
    • Example: \[ \int(\tan x)^2(\sec x)^4\,\mathrm dx=\int(\tan x)^2(1+(\tan x)^2)(\sec x)^2\,\mathrm dx=\int u^3+u^5\,\mathrm du=\frac14(\tan x)^4+\frac16(\tan x)^6+C \]
  • If \(m\) is odd.
    • You can use \((\tan x)^2=(\sec x)^2-1\) to get an expression of \(\sec x\) multiplied by \(\tan x\cdot\sec x\).
    • Since \((\sec x)'=\tan x\cdot\sec x\), you can substitute \(u=\sec x\).
    • Example: \[ \int(\tan x)^7(\sec x)^4\,\mathrm dx=\int((\sec x)^2-1)^3(\sec x)^3\tan x\cdot\sec x\,\mathrm dx=\int u^9-3u^7+3u^5-u^3\,\mathrm du=\frac{1}{10}(\sec x)^{10}-\frac38(\sec x)^8+\frac12(\sec x)^6-\frac14(\sec x)^4+C. \]

\(\int(\tan x)^m(\sec x)^n\,\mathrm dx\)

  • If neither \(m\) is odd or \(n\) is even, then there's no general algorithm.
  • \(\int\sec x\,\mathrm dx\).
    • First, we rewrite the integrand: \[ =\int\sec x\frac{\tan x+\sec x}{\tan x+\sec x}\,\mathrm dx=\int\frac{\tan x\cdot\sec x+(\sec x)^2}{\tan x+\sec x}\,\mathrm dx. \]
    • Then we substitute \(u=\tan x+\sec x\): \[ =\int\frac{1}{u}\,\mathrm du=\ln|\tan x+\sec x|+C. \]
  • \(\int(\sec x)^3\,\mathrm dx\).
    • We use integration by parts with \(u=\sec x,\,\mathrm dv=(\sec x)^2\,\mathrm dx\): \[ =\tan x\cdot\sec x-\int(\tan x)^2\sec x\,\mathrm dx. \]
    • Now we use the identity \((\tan x)^2=(\sec x)^2-1\): \[ =\tan x\cdot\sec x-\int((\sec x)^2-1)\sec x\,\mathrm dx=\tan x\cdot\sec x-\int(\sec x)^3\,\mathrm dx+\int\sec x\,\mathrm dx=\tan x\cdot\sec x-\int(\sec x)^3\,\mathrm dx+\ln|\tan x+\sec x| \]
    • This we can rearrange: \[ 2\int(\sec x)^3\,\mathrm dx=\tan x\cdot\sec x+\ln|\tan x+\sec x|+C. \]
  • Exercises. 7.2: 22, 24, 28, 30, 32

\(\int\sin mx\cdot\cos nx\,\mathrm dx\)

  • For these, we use the corresponding identity:
    • \(\sin\alpha\cos\beta=\frac12(\sin(\alpha-\beta)+\sin(\alpha+\beta))\)
    • \(\sin\alpha\sin\beta=\frac12(\cos(\alpha-\beta)-\cos(\alpha+\beta))\)
    • \(\cos\alpha\cos\beta=\frac12(\cos(\alpha-\beta)+\cos(\alpha+\beta))\).
  • For example, we have \[ \int\sin3x\cos4x\,\mathrm dx=\frac12\int\sin(-x)+\sin7x\,\mathrm dx=\frac12\cos(-x)-\frac{1}{14}\cos7x+C. \]
  • Exercises. 7.2: 41, 42, 43
  • More exercises. 7.2: 36, 40, 44, 46, 58, 62, 68