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Small note on recursive formula for integral of trigonometric functions

Nasser M. Abbasi

June 29, 2015   Compiled on May 19, 2020 at 5:46am

After stuggling in deriving this, I found similar one on wikpedia. References below. May be I will add Mathematica implementation for this later....

The goal is to find recusive formula for \(\int \cos (x)^n \,dx\). Starting by rewriting it as \begin{equation} \int \cos (x)^n \,dx = \int \cos (x)^{n-1} \cos (x) \,dx \end{equation} Integrating by parts \(\int u\,dv = (uv)-\int v\, du\) and letting \(u=\cos (x)^{n-1}, dv=\cos (x)\), hence \(du=-(n-1)\cos (x)^{n-2} \sin (x)\) and \(v=\sin (x)\) the above becomes \begin{align*} \int \cos (x)^n \,dx &= \cos (x)^{n-1} \sin (x) + \int \sin (x) (n-1)\cos (x)^{n-2} \sin (x) \,dx \\ &= \cos (x)^{n-1} \sin (x) + \int (n-1)\cos (x)^{n-2} \sin ^2(x) \,dx \\ &= \cos (x)^{n-1} \sin (x) + \int (n-1)\cos (x)^{n-2} (1-\cos (x)^2) \,dx \\ &= \cos (x)^{n-1} \sin (x) + (n-1) \int \cos (x)^{n-2} - \cos (x)^n \,dx \\ &= \cos (x)^{n-1} \sin (x) + (n-1) \int \cos (x)^{n-2}\, dx - (n-1) \int \cos (x)^n \,dx \end{align*}

The \(\int \cos (x)^n \,dx\) in the RHS above is what is being solved for. Moving it to the LHS gives \begin{align*} \int \cos (x)^n \,dx + (n-1) \int \cos (x)^n \,dx &= \cos (x)^{n-1} \sin (x) + (n-1) \int \cos (x)^{n-2}\, dx \\ n \int \cos (x)^n \,dx &= \cos (x)^{n-1} \sin (x) + (n-1) \int \cos (x)^{n-2}\, dx \end{align*}

Therefore the recusrive formula is \[ \int \cos (x)^n \,dx = \frac{\cos (x)^{n-1} \sin (x)}{n} + \frac{(n-1)}{n} \int \cos (x)^{n-2}\, dx \]

References:

  1. http://www.integraltec.com/math/math.php?f=cosPower.html\#cos
  2. http://en.wikipedia.org/wiki/Integration_by_reduction_formulae\#Examples