My Math Forum Integrals of type x^ncos(ax) or x^nsin(ax)?

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 January 19th, 2018, 03:31 AM #1 Member   Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0 Integrals of type x^ncos(ax) or x^nsin(ax)? How can you solve for these types of integrals? $\displaystyle \displaystyle \int x^{n}\cos(ax) \ dx \ or \displaystyle \int x^{n}\sin(ax) \ dx\\$ An example of an integral that I want to solve $\displaystyle \displaystyle \int x^{5}\cos(3x) \ dx$ Using u-substitution: $\displaystyle \displaystyle u = \cos(3x)\\ \displaystyle du = -3\sin(3x) \ dx\\$ Then going back to solve the integral $\displaystyle$\displaystyle \int x^{5}u \ du But how do I go on from here?
 January 19th, 2018, 04:20 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 Use integration by parts, repeatedly. To integrate $\int x^5 \cos(x) dx$, let $u= x^5$ and $dv= \cos(x)dx$. Then $du= 5x^4 dx$ and $v= \sin(x)$. So the integral becomes $x^5\sin(x)- 5\int x^4 \sin(x) dx$. Now let $u= x^4$ and $dv= \sin(x)$. Then $du= 4x^3 dx$ and $v= -\cos(x)$. So the integral becomes $x^5\sin(x)+ 5x^4\cos(x)- 20\int x^3 \cos(x)dx$. Keep doing that until the power of $x$ is reduced to 0. Thanks from DecoratorFawn82 Last edited by skipjack; January 19th, 2018 at 04:25 AM.
 January 19th, 2018, 04:40 AM #3 Senior Member   Joined: Oct 2009 Posts: 553 Thanks: 177 I have nothing of substance further to add. But I think that you can simplify a lot of the computations if you instead compute $$\int x^n e^{iax}dx$$ Thanks from greg1313
 January 19th, 2018, 06:30 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond $$\cos(ax)=\frac{e^{iax}+e^{-aix}}{2}$$ Substitute, expand and use a single u-substitution to evaluate. Thanks from Ould Youbba
 January 19th, 2018, 07:56 AM #5 Senior Member   Joined: Sep 2016 From: USA Posts: 472 Thanks: 262 Math Focus: Dynamical systems, analytic function theory, numerics Personally, I would expand as a power series and integrate term by term. However, the other methods presented here are just fine as well.
 January 19th, 2018, 09:08 AM #6 Math Team   Joined: Jul 2011 From: Texas Posts: 2,770 Thanks: 1424 Tabular integration ... $\displaystyle \int x^5 \cos(3x) \, dx = \color{red}{x^5 \cdot \dfrac{\sin(3x)}{3}} \color{blue}{+ 5x^4 \cdot \dfrac{\cos(3x)}{9}} \color{red}{- 20x^3 \cdot \dfrac{\sin(3x)}{27}} \color{blue}{ - 60x^2 \cdot \dfrac{\cos(3x)}{81}} \color{red}{+ 120x \cdot \dfrac{\sin(3x)}{243}} \color{blue}{+ 120 \cdot \dfrac{\cos(3x)}{729}} + C$ coefficients may be reduced and sine/cosine terms collected if desired ... Thanks from greg1313
January 19th, 2018, 10:30 AM   #7
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Quote:
 Originally Posted by greg1313 $$\cos(ax)=\frac{e^{iax}+e^{-aix}}{2}$$ Substitute, expand and use a single u-substitution to evaluate.
I'm not seeing this at all (the substitution).

 January 19th, 2018, 11:38 AM #8 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond $$\frac12\int x^5(e^{iax}+e^{-iax})\,dx$$ $$\frac12\int x^5e^{iax}+x^5e^{-iax}\,dx$$ $$u=e^{iax},\quad\frac{du}{ia}=e^{iax}\,dx$$ O.k., I see my error. Repeated IBP is required to integrate the resulting logarithmic expression.

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