Question

Evaluate the following integrals:

${\int }_{-\frac{3\mathrm{\pi }}{2}}^{-\frac{\mathrm{\pi }}{2}}\left\{{\sin }^2\left(3\mathrm{\pi }+x\right)+{\left(\mathrm{\pi }+x\right)}^3\right\}dx$

Answer 1

Let I = ${\int }_{-\frac{3\mathrm{\pi }}{2}}^{-\frac{\mathrm{\pi }}{2}}\left\{{\sin }^2\left(3\mathrm{\pi }+x\right)+{\left(\mathrm{\pi }+x\right)}^3\right\}dx$

Put $\mathrm{\pi }+x=z$

$\Rightarrow dx=dz$

When $x\to -\frac{3\mathrm{\pi }}{2},z\to -\frac{\mathrm{\pi }}{2}$

When $x\to -\frac{\mathrm{\pi }}{2},z\to \frac{\mathrm{\pi }}{2}$

$$\begin{gathered} \therefore I={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left[{\sin }^2\left(2\mathrm{\pi }+z\right)+z^3\right]dz \\ ={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left({\sin }^{2}z+z^3\right)dz\left[\sin \left(2\mathrm{\pi }+\theta \right)=\sin \theta \right] \\ ={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{1-\cos 2z}{2}dz+{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}{z}^{3}dz \end{gathered}$$
$$\begin{gathered} =\frac{1}{2}{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}dz-\frac{1}{2}{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\cos 2zdz+{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}{z}^{3}dz \\ =\frac{1}{2}\times {\overline{)z}}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}-\frac{1}{2}\times {\overline{)\frac{\sin 2z}{2}}}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}+{\overline{)\frac{z^4}{4}}}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \\ =\frac{1}{2}\left[\frac{\mathrm{\pi }}{2}-\left(-\frac{\mathrm{\pi }}{2}\right)\right]-\frac{1}{4}\left[\mathrm{sin\pi }-\sin \left(-\mathrm{\pi }\right)\right]+\frac{1}{4}\left(\frac{{\mathrm{\pi }}^4}{16}-\frac{{\mathrm{\pi }}^4}{16}\right) \end{gathered}$$
$$\begin{gathered} =\frac{1}{2}\times \mathrm{\pi }-\frac{1}{4}\left(0+0\right)+\frac{1}{4}\times 0 \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$