Q. 73 - Calculus

Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin x\cos xdx$

Step-by-Step Solution

Answer

The solution of the given integral is ${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin x\cos xdx=0$.

Step 1. Given Information

Solving the given integrals.

${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin x\cos xdx$

Step 2. Using the substitution method.

Let

$$\begin{gathered} u=\sin x \\ \frac{du}{dx}=\cos x \\ du=\cos xdx \end{gathered}$$

Step 3. Using the information in equations, we can change variables completely:

$$\begin{gathered} {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin x\cos xdx={\int }_{x=-\mathrm{\pi }}^{\mathrm{x}=\mathrm{\pi }}udu \\ {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin x\cos xdx={\left[\frac{u^{1+1}}{1+1}\right]}_{x=-\mathrm{\pi }}^{\mathrm{x}=\mathrm{\pi }} \\ {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin x\cos xdx={\left[\frac{u^2}{2}\right]}_{x=-\mathrm{\pi }}^{\mathrm{x}=\mathrm{\pi }} \\ {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin x\cos xdx={\left[\frac{{\sin }^{2}x}{2}\right]}_{x=-\mathrm{\pi }}^{\mathrm{x}=\mathrm{\pi }} \\ {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin x\cos xdx=\frac{1}{2}{\left[{\left(\sin x\right)}^2\right]}_{x=-\mathrm{\pi }}^{\mathrm{x}=\mathrm{\pi }} \\ {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin x\cos xdx=\frac{1}{2}\left[{\left(\mathrm{sin\pi }\right)}^2-{\left\{\sin \right(-\pi \left)\right\}}^2\right] \\ {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin x\cos xdx=\frac{1}{2}\left[{\left(\mathrm{sin\pi }\right)}^2-{\{-\sin (\pi )\}}^2\right] \\ {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin x\cos xdx=\frac{1}{2}\left[{\left(0\right)}^2-{\left(0\right)}^2\right] \\ {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin x\cos xdx=0 \end{gathered}$$