Question

how to integrate this?​

Answer 1

Given to find,

$\displaystyle\int\limits_0^{\frac{3}{2}}x\sin (\pi x)\ dx$

Recall the product rule of differentiation.

$\displaystyle\int uv'=uv-\int u'v$

Thus, let,

$\begin {aligned}&u=x\\\\\implies\ \ &u'=1\end {aligned}$

and,

$\displaystyle\begin {aligned}&v'=\sin (\pi x)\\\\\\\implies\ \ &v=\int\sin (\pi x) dx\\\\\\&v=\dfrac {1}{\pi}\int\sin (\pi x)\ d(\pi x)\\\\\\&v=-\dfrac {\cos (\pi x)}{\pi}\end{aligned}$

Then,

$\displaystyle\int\limits_0^{\frac{3}{2}}x\sin (\pi x)\ dx\\\\\\=\left [-\dfrac {x\cos (\pi x)}{\pi}\right]_0^\frac {3}{2}+\dfrac {1}{\pi}\int\limits_0^\frac {3}{2}\cos(\pi x)\ dx\\\\\\=-\dfrac {1}{\pi}\left [x\cos (\pi x)\right]_0^\frac {3}{2}+\dfrac {1}{\pi^2}\int\limits_0^\frac {3}{2}\cos(\pi x)\ d(\pi x)\\\\\\=-\dfrac {1}{\pi}\left [x\cos (\pi x)\right]_0^\frac {3}{2}+\dfrac {1}{\pi^2}\left [\sin (\pi x)\right]_0^{\frac {3}{2}}$

$=-\dfrac {1}{\pi}\left [\left (\dfrac {3}{2}\cos \left(\dfrac {3\pi}{2}\right)\right)-\left (0\cos (\pi \cdot 0)\right)\right]+\dfrac {1}{\pi^2}\left [\sin \left(\dfrac {3\pi}{2}\right)-\sin (\pi\cdot 0)\right]\\\\\\=-\dfrac {1}{\pi}\left [\left (\dfrac {3}{2}\cdot 0\right)-\left (0\cdot1\right)\right]+\dfrac {1}{\pi^2}\left (-1-0\right)\\\\\\=\large\boxed {\mathbf{-\dfrac {1}{\pi^2}}}$

Hence integrated!