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## Small note on recursive formula for integral of trigonometric functions

June 29, 2015   Compiled on May 26, 2022 at 4:57pm

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 $\int \cos (x)^n \,dx = \frac {\cos (x)^{n-1} \sin (x)}{n} + \frac {(n-1)}{n} \int \cos (x)^{n-2}\, dx$

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