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January 19th, 2018, 04:31 AM   #1
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Question 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?
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January 19th, 2018, 05:20 AM   #2
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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.
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Last edited by skipjack; January 19th, 2018 at 05:25 AM.
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January 19th, 2018, 05:40 AM   #3
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I have nothing of substance further to add. But I think that you can simplify a lot of the computations if you instead compute

$$\int x^n e^{iax}dx$$
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January 19th, 2018, 07:30 AM   #4
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$$\cos(ax)=\frac{e^{iax}+e^{-aix}}{2}$$

Substitute, expand and use a single u-substitution to evaluate.
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January 19th, 2018, 08:56 AM   #5
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Personally, I would expand as a power series and integrate term by term. However, the other methods presented here are just fine as well.
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January 19th, 2018, 10:08 AM   #6
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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 ...
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January 19th, 2018, 11:30 AM   #7
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Quote:
Originally Posted by greg1313 View Post
$$\cos(ax)=\frac{e^{iax}+e^{-aix}}{2}$$

Substitute, expand and use a single u-substitution to evaluate.
I'm not seeing this at all (the substitution).
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January 19th, 2018, 12:38 PM   #8
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$$\frac12\int x^5(e^{iax}+e^{-iax})\,dx$$

$$\frac12\int x^5e^{iax}+x^5e^{-iax}\,dx$$

$$u=e^{iax},\quad\frac{du}{ia}=e^{iax}\,dx$$

O.k., I see my error. Repeated IBP is required to integrate the resulting logarithmic expression.
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