Question

Evaluating the integral

\(\smallint \left( {{\bf{x}} - 1} \right)sin\pi x\;dx\)

Short answer

We should useintegration by parts to solve the given integral

Let, u = x-1 and dv = sinπx

Evaluate du and v, and substitute it

\(\smallint udv\; = \;uv\; - \;\smallint vdu\)

Step-by-step solution

- Given Data

Given = \(\)\(\)\(\smallint \left( {x - } \right)sin\pi x{\rm{\;}}dx\)

Let, u = x-1 \(\smallint dv\; = \;\smallint sin\pi x{\rm{\;}}dx\)

du=dx \(v\; = \; - \frac{{cos\pi x}}{\pi }\)

-Integrating the equation

\(\begin{array}{l}\int {(x - 1)\sin \pi xdx = - (x - 1) - \int { - \frac{{\cos \pi x}}{\pi }} } dx\\ = - (x - 1)\frac{{\cos \pi x}}{\pi } + \frac{{\sin \pi x}}{{\pi (\pi )}} + c\\ = - \frac{1}{x}\left( {(x - 1)\cos \pi x - \frac{{\sin \pi x}}{\pi }} \right) + c\end{array}\)

Hence,\(\int {(x - 1)\sin \pi xdx = - \frac{1}{x}\left( {(x - 1)\cos \pi x - \frac{{\sin \pi x}}{\pi }} \right) + c} \)