Question

Evaluate the following integrals:

${\int }_{-1}^1\left|x\mathrm{cos\pi }x\right|dx$ [NCERT EXEMPLAR]

Answer 1

Let I = ${\int }_{-1}^1\left|x\mathrm{cos\pi }x\right|dx$

Consider $f\left(x\right)=\left|x\mathrm{cos\pi }x\right|$.

$f\left(-x\right)=\left|\left(-x\right)\mathrm{cos\pi }\left(-x\right)\right|=\left|-x\mathrm{cos\pi }x\right|=\left|x\mathrm{cos\pi }x\right|=f\left(x\right)$

$$\begin{gathered} \therefore I={\int }_{-1}^1\left|x\mathrm{cos\pi }x\right|dx \\ =2{\int }_0^1\left|x\mathrm{cos\pi }x\right|dx\left[{\int }_{-a}^{a}f\left(x\right)dx=\left\{\begin{array}{ll}2{\int }_0^{a}f\left(x\right)dx,& \mathrm{if}f\left(-x\right)=f\left(x\right)\\ 0,& \mathrm{if}f\left(-x\right)=-f\left(x\right)\end{array}\right.\right] \end{gathered}$$
Now,

$\left|x\mathrm{cos\pi }x\right|=\left\{\begin{array}{ll}x\mathrm{cos\pi }x,& \mathrm{if}0\le x\le \frac{1}{2}\\ -x\mathrm{cos\pi }x,& \mathrm{if}\frac{1}{2}<x\le 1\end{array}\right.$

$$\begin{gathered} \therefore I=2\left[{\int }_0^{\frac{1}{2}}x\mathrm{cos\pi }xdx+{\int }_{\frac{1}{2}}^1\left(-x\mathrm{cos\pi }x\right)dx\right] \\ =2\left[x{\overline{)\frac{\mathrm{sin\pi }x}{\mathrm{\pi }}}}_0^{\frac{1}{2}}-\frac{1}{\mathrm{\pi }}{\int }_0^{\frac{1}{2}}\mathrm{sin\pi }xdx\right]-2\left[x{\overline{)\frac{\mathrm{sin\pi }x}{\mathrm{\pi }}}}_{\frac{1}{2}}^1-\frac{1}{\mathrm{\pi }}{\int }_{\frac{1}{2}}^1\mathrm{sin\pi }xdx\right] \\ =2\left(\frac{1}{2\mathrm{\pi }}\sin \frac{\mathrm{\pi }}{2}-0\right)-\frac{2}{\mathrm{\pi }}\times {\overline{)\left(-\frac{\mathrm{cos\pi x}}{\mathrm{\pi }}\right)}}_0^{\frac{1}{2}}-2\left(\frac{1}{\mathrm{\pi }}\mathrm{sin\pi }-\frac{1}{2\mathrm{\pi }}\sin \frac{\mathrm{\pi }}{2}\right)+\frac{2}{\mathrm{\pi }}\times {\overline{)\left(-\frac{\mathrm{cos\pi }x}{\mathrm{\pi }}\right)}}_{\frac{1}{2}}^1 \\ =\frac{1}{\mathrm{\pi }}+\frac{2}{{\mathrm{\pi }}^2}\left(\cos \frac{\mathrm{\pi }}{2}-\cos 0\right)+\frac{1}{\mathrm{\pi }}-\frac{2}{{\mathrm{\pi }}^2}\left(\mathrm{cos\pi }-\cos \frac{\mathrm{\pi }}{2}\right) \\ =\frac{1}{\mathrm{\pi }}-\frac{2}{{\mathrm{\pi }}^2}+\frac{1}{\mathrm{\pi }}+\frac{2}{{\mathrm{\pi }}^2} \\ =\frac{2}{\mathrm{\pi }} \end{gathered}$$