Question

Evaluate the following integral:

\[\int_{- 1}^1 \left| xcos\pi x \right|dx\]

 

Answer 1

\[\text{Let I} =\int_{- 1}^1 \left| xcos\pi x \right|dx\]

Consider

\[f\left( x \right) = \left| xcos\pi x \right|\]

\[f\left( - x \right) = \left| \left( - x \right)cos\pi\left( - x \right) \right| = \left| - xcos\pi x \right| = \left| xcos\pi x \right| = f\left( x \right)\]

\[\therefore I = \int_{- 1}^1 \left| xcos\pi x \right|dx\]
\[ = 2 \int_0^1 \left| xcos\pi x \right|dx ...............\left[ \int_{- a}^a f\left( x \right)dx = \begin{cases}2 \int_0^a f\left( x \right)dx, & \text{if }f\left( - x \right) = f\left( x \right) \\ 0, & \text{if }f\left( - x \right) = - f\left( x \right)\end{cases} \right]\]

Now,

\[\left| xcos\pi x \right| = \begin{cases}xcos\pi x, & \text{if  }0 \leq x \leq \frac{1}{2} \\ - xcos\pi x, & \text{if }\frac{1}{2} < x \leq 1\end{cases}\]

\[\therefore I = 2\left[ \int_0^\frac{1}{2} xcos\pi xdx + \int_\frac{1}{2}^1 \left( - xcos\pi x \right)dx \right]\]
\[ = \left.2\left[ x \frac{sin\pi x}{\pi}\right|_0^\frac{1}{2} -\left. \frac{1}{\pi} \int_0^\frac{1}{2} sin\pi xdx \right] - 2\left[ x \frac{sin\pi x}{\pi}\right|_\frac{1}{2}^1 - \frac{1}{\pi} \int_\frac{1}{2}^1 sin\pi xdx \right]\]
\[ = 2\left( \frac{1}{2\pi}\sin\frac{\pi}{2} - 0 \right) - \left.\frac{2}{\pi} \times \left( - \frac{cos\pi x}{\pi} \right)\right|_0^\frac{1}{2} - \left.2\left( \frac{1}{\pi}sin\pi - \frac{1}{2\pi}\sin\frac{\pi}{2} \right) + \frac{2}{\pi} \times \left( - \frac{cos\pi x}{\pi} \right)\right|_\frac{1}{2}^1 \]
\[ = \frac{1}{\pi} + \frac{2}{\pi^2}\left( \cos\frac{\pi}{2} - \cos0 \right) + \frac{1}{\pi} - \frac{2}{\pi^2}\left( cos\pi - \cos\frac{\pi}{2} \right)\]
\[ = \frac{1}{\pi} - \frac{2}{\pi^2} + \frac{1}{\pi} + \frac{2}{\pi^2}\]
\[ = \frac{2}{\pi}\]