evaluate the integral int x^2 sin pi x dx

[sy211006]

evaluate the integral int x^2 sin pi xdx
ok so i let u=x^2 and dv=sin pi x
-x^2 1/pi cos pie x + 1/pie int. cos pi x (2x)
-x^2 1/pie cos pi x + 2/pie int. x cos pi x
then i let u=x dv=cos pi x
-x^2 1/pi cos pi x + 2/pi ^2 x sin pi x + 1/pie int. sin pi x
so again i let u=x and dv=sin pi
here is where i am stuck!!

the answer in the book is -1/pi x^2 cos pi x + 2/ pi ^2 x sin pi x + 2/pi ^3 cos pi x + c
i got the first two terms but am having problems with the last term
Please Help!!

 

[royhaas]

Re: help i just can figure this out

Your last integral should be \(\displaystyle -\frac{2}{\pi^2}\int \sin(\pi x)dx\).

 

[sy211006]

Re: help i just can figure this out

How did u get that

 

[Deleted member 4993]

Re: help i just can figure this out

evaluate the integral int x^2 sin pie xdx
ok so i let u=x^2 and dv=sin pie x
-x^2 1/pie cos pie x + 1/pie int. cos pie x (2x)
-x^2 1/pie cos pie x + 2/pie int. x cos pie x
then i let u=x dv=cos pie x
-x^2 1/pie cos pie x + 2/pie ^2 x sin pie x + 1/pie int. sin pie x <<< This is incorrect
so again i let u=x and dv=sin pie
here is where i am stuck!!

the answer in the book is -1/pie x^2 cos pie x + 2/ pie ^2 x sin pie x + 2/pie ^3 cos pie x + c
i got the first two terms but am having problems with the last term
Please Help!!

 

[soroban]

Re: evaluate the integral int x^2 sin pie x dx

Hello, sy211006!

Okay, let's take it from the top . . .
(By the way, it's "pi", not a crusted dessert.)

\(\displaystyle I \;=\;\int x^2\sin(\pi x)\,dx\)

. . \(\displaystyle \begin{array}{ccccccc}u &=& x^2 & & dv &=&\sin(\pi x)\,dx \\ dy &=& 2x\,dx & & v &=&\text{-}\frac{1}{\pi}\cos(\pi x) \end{array}\)

\(\displaystyle \text{We have: }\;I \;=\;-\frac{1}{\pi}\,x^2\cos(\pi x) + \frac{2}{\pi} \int x\cos(\pi x)\,dx\)


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

\(\displaystyle \text{We have: }\;I \;=\;-\frac{1}{\pi}\,x^2\cos(\pi x) + \frac{2}{\pi}\bigg[\frac{1}{\pi}x\sin(\pi x) - \frac{1}{\pi}\int\sin(\pi x)\,dx\bigg]\)

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

. . . . . . . . .\(\displaystyle = \;-\frac{1}{\pi}\,x^2\cos(\pi x) + \frac{2}{\pi^2}\,x\sin(\pi x) + \frac{2}{\pi^3}\cos(\pi x) + C\)

 

[k082191]

Hello, sy211006!

Okay, let's take it from the top . . .
(By the way, it's "pi", not a crusted dessert.)

. . \(\displaystyle \begin{array}{ccccccc}u &=& x^2 & & dv &=&\sin(\pi x)\,dx \\ dy &=& 2x\,dx & & v &=&\text{-}\frac{1}{\pi}\cos(\pi x) \end{array}\)

\(\displaystyle \text{We have: }\;I \;=\;-\frac{1}{\pi}\,x^2\cos(\pi x) + \frac{2}{\pi} \int x\cos(\pi x)\,dx\)


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

\(\displaystyle \text{We have: }\;I \;=\;-\frac{1}{\pi}\,x^2\cos(\pi x) + \frac{2}{\pi}\bigg[\frac{1}{\pi}x\sin(\pi x) - \frac{1}{\pi}\int\sin(\pi x)\,dx\bigg]\)

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

. . . . . . . . .\(\displaystyle = \;-\frac{1}{\pi}\,x^2\cos(\pi x) + \frac{2}{\pi^2}\,x\sin(\pi x) + \frac{2}{\pi^3}\cos(\pi x) + C\)

Pardon me I had to make the font smaller to see it.