Selena's question at Yahoo Answers regarding a definite integral by parts

[MarkFL]

Here is the question:

Urgent Integral by parts?


1/pi ∫[limits from 0 to 1] (1-v^2)cos(wv) dv

The answer is 1/(pi*w) [(2/w^2)*sinw - (2/w)*cosw]

Please explain in detail. Thank you.

I have posted a link there to this thread so the OP can see my work.

 

[MarkFL]

Hello Selena,

We are given to evaluate:

$$I=\frac{1}{\pi}\int_0^1\left(1-v^2 \right)\cos(wv)\,dv$$

Using integration by parts, let's let:

$$s=1-v^2\,\therefore\,ds=-2v\,dv$$

$$dt=\cos(wv)\,dv\,\therefore\,t=\frac{1}{w}\sin(wv)$$

Hence:

$$I=\frac{1}{\pi}\left(\left[\frac{1-v^2}{w}\sin(wv) \right]_0^1+\frac{2}{w}\int_0^1 v\sin(wv)\,dv \right)$$

$$I=\frac{2}{\pi w}\int_0^1 v\sin(wv)\,dv$$

Using integration by parts again, let's let:

$$s=v\,\therefore\,ds=dv$$

$$dt=\sin(wv)\,dv\,\therefore\,t=-\frac{1}{w}\cos(wv)$$

Hence:

$$I=\frac{2}{\pi w}\left(\left[-\frac{v}{w}\cos(wv) \right]_0^1+\frac{1}{w}\int_0^1 \cos(wv)\,dv \right)$$

$$I=\frac{2}{\pi w}\left(-\frac{1}{w}\cos(w)+\frac{1}{w^2}\left[\sin(wv) \right]_0^1 \right)$$

$$I=\frac{1}{\pi w}\left(\frac{2}{w^2}\sin(w)-\frac{2}{w}\cos(w) \right)$$