integration by parts: integral of x^2 cos mx dx

[xc630]

Hello I would appreciate any help with the following problem:

The integral of x^2 cos mx dx

I tried letting u=x^2 and dv= cos mx dx and got to x^2 (1/m) sin mx - (1/m)* the integral of sin mx 2x dx. From here I tried applying integration by parts again but haven't gotten anywhere

 

[soroban]

Re: integration by parts

Hello, xc630!

\(\displaystyle \int x^2\cos^2(mx)\,dx\)

\(\displaystyle \text{Your first by-parts is correct . . .}\)

. . \(\displaystyle \begin{array}{ccccccc}u & = & x^2 & \quad & dv & = & \cos(mx)\,dx \\ du & = & 2x\,dx & \quad & v & = & \frac{1}{m}\sin(mx) \end{array}\)

\(\displaystyle \text{We have: }\;\frac{1}{m}x^2\sin(mx) - \frac{2}{m}\int x\sin(mx)\,dx\)


\(\displaystyle \text{Then: }\;\begin{array}{ccccccc}u & = & x & \quad & dv & = & \sin(mx)\,dx \\ du &=& dx & \quad & v &=&\text{-}\frac{1}{m}\cos(mx) \end{array}\)

\(\displaystyle \text{We have: }\;\frac{1}{m}x^2\sin(mx) - \frac{2}{m}\left[-\frac{1}{m}x\cos(mx) + \frac{1}{m}\int\cos(mx)\,dx\right]\)

. . . . . . . \(\displaystyle =\;\frac{1}{m}x^2\sin(mx) + \frac{2}{m^2}x\cos(mx) - \frac{2}{m^2}\int\cos(mx)\,dx\)


Can you finish it now?

 

[xc630]

Re: integration by parts

thanks for the help soroban

 

[Deleted member 4993]

Re: integration by parts

Hello, xc630!

\(\displaystyle \int x^2\cos^2(mx)\,dx\)

\(\displaystyle \text{Your first by-parts is correct . . .}\)

. . \(\displaystyle \begin{array}{ccccccc}u & = & x^2 & \quad & dv & = & \cos(mx)\,dx \\ du & = & 2x\,dx & \quad & v & = & \frac{1}{m}\sin(mx) \end{array}\)

Isn't it \(\displaystyle cos^2(mx)\)

\(\displaystyle \text{We have: }\;\frac{1}{m}x^2\sin(mx) - \frac{2}{m}\int x\sin(mx)\,dx\)


\(\displaystyle \text{Then: }\;\begin{array}{ccccccc}u & = & x & \quad & dv & = & \sin(mx)\,dx \\ du &=& dx & \quad & v &=&\text{-}\frac{1}{m}\cos(mx) \end{array}\)

\(\displaystyle \text{We have: }\;\frac{1}{m}x^2\sin(mx) - \frac{2}{m}\left[-\frac{1}{m}x\cos(mx) + \frac{1}{m}\int\cos(mx)\,dx\right]\)

. . . . . . . \(\displaystyle =\;\frac{1}{m}x^2\sin(mx) + \frac{2}{m^2}x\cos(mx) - \frac{2}{m^2}\int\cos(mx)\,dx\)


Can you finish it now?

[/size]

 

[Deleted member 4993]

Re: integration by parts

Hello, xc630!

\(\displaystyle \int x^2\cos^2(mx)\,dx\)

\(\displaystyle cos^2(mx) = \frac {1 + cos(2mx)}{2}\cdot dx\)

\(\displaystyle x^2\cos^2(mx)\cdot dx = \frac {x^2}{2}\cdot dx +\frac{1}{16m^3}\cdot (2mx)^2cos(2mx)\cdot d(2mx)\)

\(\displaystyle x^2\cos^2(mx)\cdot dx = \frac {x^2}{2}\cdot dx +\frac{1}{16m^3}\cdot w^2cos(w)\cdot dw\)

Now follow Soroban's method described below......



\(\displaystyle \text{Your first by-parts is correct . . .}\)

. . \(\displaystyle \begin{array}{ccccccc}u & = & x^2 & \quad & dv & = & \cos(mx)\,dx \\ du & = & 2x\,dx & \quad & v & = & \frac{1}{m}\sin(mx) \end{array}\)

\(\displaystyle \text{We have: }\;\frac{1}{m}x^2\sin(mx) - \frac{2}{m}\int x\sin(mx)\,dx\)


\(\displaystyle \text{Then: }\;\begin{array}{ccccccc}u & = & x & \quad & dv & = & \sin(mx)\,dx \\ du &=& dx & \quad & v &=&\text{-}\frac{1}{m}\cos(mx) \end{array}\)

\(\displaystyle \text{We have: }\;\frac{1}{m}x^2\sin(mx) - \frac{2}{m}\left[-\frac{1}{m}x\cos(mx) + \frac{1}{m}\int\cos(mx)\,dx\right]\)

. . . . . . . \(\displaystyle =\;\frac{1}{m}x^2\sin(mx) + \frac{2}{m^2}x\cos(mx) - \frac{2}{m^2}\int\cos(mx)\,dx\)


Can you finish it now?

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