Question

Evaluate the following integrals as they are written. $${\int }_0^{\pi /2}{\int }_y^{\pi /2}6\sin (2x-3y)dxdy$$

Short answer

Question: Evaluate the double integral of the function $${\int }_0^{\pi /2}{\int }_y^{\pi /2}6\sin (2x-3y)dxdy$$. Answer: The double integral evaluates to 2.

Step-by-step solution

Integrate with respect to x

First, we need to integrate the function with respect to x: $${\int }_0^{\pi /2}{\int }_y^{\pi /2}6\sin (2x-3y)dxdy$$ We will integrate the inner integral by treating y as a constant. The integral of a function containing sine is -cosine, with the argument remaining the same. However, if the coefficient of x inside the argument is not 1, we must divide the result by that coefficient. In this case, the coefficient is 2. $$\Rightarrow {\int }_0^{\pi /2}-3{\left[\frac{\cos (2x-3y)}{2}\right]}_{x=y}^{x=\pi /2}dy$$

Plug in the x limits

Now, we need to plug in the x limits in the integral expression: $$\Rightarrow {\int }_0^{\pi /2}-3[\frac{\cos (2(\pi /2)-3y)}{2}-\frac{\cos (2y-3y)}{2}]dy$$ Simplify the expression inside the integral: $$\Rightarrow -\frac{3}{2}{\int }_0^{\pi /2}[\cos (\pi -3y)-\cos (-y)]dy$$ Since cosine is an even function, $\cos (-y)=\cos (y)$, the expression becomes: $$\Rightarrow -\frac{3}{2}{\int }_0^{\pi /2}[\cos (\pi -3y)-\cos (y)]dy$$

Integrate with respect to y

Now we'll integrate the expression with respect to y: $$\Rightarrow -\frac{3}{2}[{\int }_0^{\pi /2}\cos (\pi -3y)dy-{\int }_0^{\pi /2}\cos (y)dy]$$ Integrating the cosine function with respect to y gives us a sine function, and we need to divide by the coefficient of y inside the argument, just like we did with x: $$\Rightarrow -\frac{3}{2}{[\frac{\sin (\pi -3y)}{3}-\sin (y)]}_{y=0}^{y=\pi /2}$$

Plug in the y limits

Finally, we plug in the y limits of the integral: $$\Rightarrow -\frac{3}{2}[\frac{\sin (\pi -3(\pi /2))}{3}-\sin (\pi /2)-\frac{\sin (\pi -3(0))}{3}+\sin (0)]$$ Calculate the result: $$\Rightarrow -\frac{3}{2}[\frac{\sin (-\pi /2)}{3}-1-\frac{\sin (\pi )}{3}+0]$$ The values of sine are: $$\Rightarrow -\frac{3}{2}[\frac{-1}{3}-1-0+0]$$ Solve the final expression: $$\Rightarrow -\frac{3}{2}\left(\frac{-4}{3}\right)=\boxed{{\displaystyle 2}}$$

Key concepts

Integrating Trigonometric Functions

Integrating trigonometric functions is a fundamental technique often encountered in calculus. When dealing with integrals that involve trigonometric expressions, understanding the periodic properties of these functions such as sine and cosine is crucial. For instance, in the exercise we consider the definite integral involving the function $6\sin (2x-3y)$.

To integrate this, one must recall that the integral of sine is negative cosine, but one must be careful when the argument of the sine function includes a coefficient other than 1. Here, the coefficient is 2 in the argument $2x-3y$, demanding adjustment by the reciprocal of this coefficient after integration. Hence, $\int \sin (ax)dx=-\frac{1}{a}\cos (ax)+C$, where $C$ is the constant of integration. Further, cosines' even nature, i.e., $\cos (-x)=\cos (x)$, simplifies the expression when substituting the limits of integration.

Definite Integrals

Definite integrals, unlike indefinite integrals, provide the net area between the function and the x-axis over a specific interval. To solve the definite integral, one usually finds the indefinite integral first, and then applies the Fundamental Theorem of Calculus by subtracting the value of the antiderivative at the lower limit from the value at the upper limit.

In our exercise, after integrating with respect to $x$, the next step involves evaluating the resulting expression from $y=0$ to $y=\frac{\pi }{2}$. Care must be taken to evaluate the sine and cosine functions correctly at their respective limits. For instance, $\cos (\pi -3y)$ and $\sin (\pi -3y)$ require attention to the values of trigonometric functions at special angles to determine the integral's value at the bounds. It's essential to recognize that $\sin (\pi )=0$ and $\sin (\frac{-\pi }{2})=-1$, to correctly compute the final answer.

Integration Techniques

The vast array of integration techniques is what makes calculating integrals feasible in various situations. Key strategies include substitution, integration by parts, partial fractions, and trigonometric identities, each with its own set of guidelines for application.

When applying these techniques to solve double integrals, as seen in the provided exercise, you must integrate with respect to one variable while considering the other as a constant. This approach transforms a complex double integral into a step-by-step process where we tackle one integral at a time. In some cases, recognizing an integral as a standard form can save time and reduce errors in computation. For example, knowing that inverses of derivatives of trigonometric function—such as $-\frac{1}{a}\cos (ax)$ being the integral of $\sin (ax)$—allows for quick integration.

As part of improving the understanding of this process, it is helpful to draw attention to the importance of evaluating the resulting functions at the limits of integration, paying close attention to the signs and the values of trigonometric functions at specific angles. In this way, students can better comprehend the connections between the steps and the overall structure of double integration problems.