Chapter 14, Problem 1RE

Evaluating an Integral In Exercises 1 and 2, evaluate the integral.

${\displaystyle {\int }_0^{3x}\sin (xy)dy}$

To determine

To calculate: The value of the integral given as ${\displaystyle {\int }_0^{3x}\sin \left(xy\right)dy}$.

Answer to Problem 1RE

Solution: $\frac{1-\cos \left(3x^2\right)}{x}$

Explanation of Solution

Given: The provided integral is ${\displaystyle {\int }_y^{y^2}\frac{x}{y+1}dx}$.

Formula used: The integral property of trigonometry given as:

${\displaystyle \int \sin \left(ax\right)}dx=\frac{-\cos \left(ax\right)}{a}$

Calculation: The function is integrated with respect to y, taking x as a constant variable:

$\begin{array}{c}{\displaystyle {\int }_0^{3x}\sin \left(xy\right)dy}={\left[\frac{-\cos \left(xy\right)}{x}\right]}_0{}^{3x}\\ =\frac{-\cos \left(x\cdot \left(3x\right)\right)}{x}-\left(\frac{-\cos \left(x\cdot 0\right)}{x}\right)\\ =\frac{-\cos \left(3x^2\right)}{x}+\frac{1}{x}\\ =\frac{1-\cos \left(3x^2\right)}{x}\end{array}$