# Integration By Parts

• Mar 20th 2013, 08:46 PM
knag95
Integration By Parts
Could someone please do me a favor and show me the steps to integrate this problem: 2PI fnInt((x)(-10cos((PI/6.44)x)+18 - 1 ),x,0,0.2)? Thanks.
• Mar 21st 2013, 02:36 AM
chiro
Re: Integration By Parts
Hey knag95.

It is usually a better idea for the person attempting the question to show what they have tried rather than to just be handed the answer. Can you show us what you tried?
• Mar 30th 2013, 02:51 PM
x3bnm
Re: Integration By Parts
We can simplify to write:

\begin{align*}2*\pi \int_{0}^{0.2} x * \left(-10 \cos \left(\frac{\pi * x}{6.44}\right) + 18 -1\right) \,\,dx = 2 * \pi \int_{0}^{0.2} x * & \Big(-10 \cos \left(\frac{\pi * x}{6.44}\right)\\ &+ 17\Big)\,\,dx\end{align*}

Now we know that if $u(x)$ and $v(x)$ are two functions then:

$\int \frac{d(u(x)v(x))}{dx} dx = u(x)v(x) = \int u(x) dv + \int v(x) du$

If we rearrange this we get:

$\int u(x) dv = u(x)v(x) - \int v(x) du.......\text{Equation(1)}$

Let $u(x) = x \text{ then } \frac{du}{dx} = 1 \;\;\therefore du = dx$ and let:

$dv = \left(-10 \cos \left(\frac{\pi * x}{6.44}\right) + 17\right)\,\,dx$

then by integrating both sides:

$v(x) = \frac{-10 * 6.44 \sin \left(\frac{\pi * x}{6.44}\right)}{\pi}+ 17x$

Now by plugging in appropriate values to Equation(1) we get:

\begin{align*}& 2*\pi \int_{0}^{0.2} x * (-10 \cos (\frac{\pi * x}{6.44}) + 17) \,\,dx\\ =& 2*\pi\Big[x * \left(\frac{-10 * 6.44 \sin (\frac{\pi * x}{6.44})}{\pi}+ 17x \right) - \int_{0}^{0.2} \left(\frac{-10 * 6.44 \sin (\frac{\pi * x}{6.44})}{\pi}+ 17x\right)\,\,dx\Big]\\=& 2*\pi \Big[ x * \left(\frac{-10 * 6.44 \sin (\frac{\pi * x}{6.44})}{\pi}+ 17x \right) - \left(\frac{10 * (6.44)^2 \cos (\frac{\pi * x}{6.44})}{(\pi)^2}+ \frac{17x^2}{2} \right)\Big]_{0}^{0.2}\Big]\\=& 2* \pi\Big[0.2 * \left(\frac{-10 * 6.44 \sin (\frac{\pi * 0.2}{6.44})}{\pi}+ (17*0.2) \right) - \left(\frac{10 * (6.44)^2 \cos (\frac{\pi * 0.2}{6.44})}{(\pi)^2}+ \frac{17*0.2^2}{2} \right)\\& - 0 + \left(\frac{10 * (6.44)^2}{(\pi)^2}+ 0\right)\Big]\\=& 2*\pi*(0.280634-42.161701+42.021542)\\=& 0.88263 \end{align*}

That's the answer. Hope it helps.

You can check the answer at: integrate (2*pi * ((x)(-10 * cos((pi/6.44) * x) +17 ))) dx from x=0 to 0.2 - Wolfram|Alpha