One function is in the form x to the power of n and the other can be integrated at least n times

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Here we can set u= xn, and then after one partial integration, we get an integrand with a factor xn. We may then repeat the process until we get to 1. The example on the previous page is trivialy is an example of this, so let us look at a somewhat more complex example.

\int {{x^2}\sin (x)dx}  

\left| {\begin{array}{*{20}{c}}  {u = {x^2}} \\   {\frac{{dv}}{{dx}} = \sin x} \\  {\frac{{du}}{{dx}} = 2x} \\   {v =  - \cos x} \end{array}} \right.

\begin{gathered}   \int {{x^2}\sin (x)dx}  =  - {x^2}\cos x - \int {2x( - \cos x)} \;dx \hfill \\   \quad  =  - {x^2}\cos x + 2\int {x\cos x\;dx}  \hfill \\  \end{gathered}

Or, written in another way,

\displaystyle \begin{array}{l}\int{{\underbrace{{{{x}^{2}}}}_{g}\underbrace{{\sin x}}_{f}dx}}=\underbrace{{(-\cos x)}}_{F}\underbrace{{{{x}^{2}}}}_{g}-\int{{\underbrace{{(-\cos x)}}_{F}}}\cdot \underbrace{{2x}}_{{{g}'}}dx\\\quad =-{{x}^{2}}\cos x+2\int{{x\cos x\,dx}}\end{array}

Then we can do an integration by parts again.

\left| {\begin{array}{*{20}{c}}   {u = x} \\    {\frac{{dv}}{{dx}} = \cos x} \\    {\frac{{du}}{{dx}} = 1} \\    {v = \sin x}  \end{array}} \right.

\begin{gathered}   - {x^2}\cos x + 2\int {x\cos x\;dx}  =  \hfill \\   \quad  =  - {x^2}\cos x + 2(x\sin x - \int {1\sin x\,dx} ) \hfill \\  \quad  =  - {x^2}\cos x + 2(x\sin x + \cos x) + C \hfill \\   \quad  =  - {x^2}\cos x + 2x\sin x + 2\cos x + C \hfill \\  \end{gathered}  

or,

\displaystyle \begin{array}{l}\int{{\underbrace{x}_{g}\underbrace{{\cos x}}_{f}\,dx}}=\underbrace{x}_{g}\underbrace{{\sin x}}_{F}-\int{{\underbrace{1}_{{{g}'}}\cdot }}\underbrace{{\sin x}}_{F}dx=\\\quad =x\sin x--\cos x+C=x\sin x+\cos x+C\end{array}

that we may substitute into the previous integral to get the same result as before.

 

Up a level : Integrals
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Next page : One of the functions is sin(bx) or cos(bx) or a linear combination of them, the other is an exponential functionLast modified: May 8, 2024 @ 12:37