Problem 12 - Calculus Early Transcendentals

Question

Evaluate the following iterated integrals. $$\int_{0}^{\pi / 2} \int_{0}^{1} x \cos x y d y d x$$

Step-by-Step Solution

Answer

Question: Evaluate the iterated integral $$\int_{0}^{\pi/2} \int_{0}^{1} x \cos{xy} d y d x$$ Answer: The iterated integral is equivalent to the integral $$\int \frac{u}{\arcsin{u}} \frac{1}{\sqrt{1 - u^2}} du$$, which cannot be directly solved in its current form.

Step 1: Evaluate the inner integral with respect to \(y\)

The inner integral is with respect to \(y\). Treat \(x\) as a constant, since it does not affect the current integration process. First, integrate the expression \(x\cos{xy}\) with respect to \(y\): $$\int_{0}^{1} x\cos{xy}dy$$ Integration: $$\int x\cos{xy} dy = \frac{1}{x}\int (\cos{xy}\cdot x) dy = \frac{1}{x}\sin{xy} + C_1(x)$$ Now, evaluate the limits for integration after applying the antiderivative: $$\left[\frac{1}{x}\sin{xy}\right]_{y=0}^{y=1} = \frac{1}{x}\sin{x} - \frac{1}{x}\sin{0} = \frac{1}{x}\sin{x}$$ Now, the expression is reduced to: $$\int_{0}^{\pi/2}\frac{1}{x}\sin{x} dx$$

Step 2: Evaluate the outer integral with respect to \(x\)

Now, we need to evaluate the outer integral with respect to \(x\): $$\int_{0}^{\pi/2}\frac{1}{x}\sin{x} dx$$ Integration by substitution can be used here. Let \(u = \sin{x}\) and \(du = \cos{x}dx\). The new integration becomes: $$\int \frac{u}{x} du$$ But we need to express \(x\) in terms of \(u\). Recall that \(u = \sin{x}\). Therefore, we can use the arcsine function: $$ x = \arcsin{u} \Rightarrow dx = \frac{1}{\sqrt{1-u^2}} du$$ The new integral turns into: $$ \int \frac{u}{\arcsin{u}} \frac{1}{\sqrt{1 - u^2}} du$$ However, this integral cannot be directly solved as it is so we leave it in the given form. Therefore, the final answer for the given iterated integral is: $$\int_{0}^{\pi/2} \int_{0}^{1} x \cos x y d y d x = \int \frac{u}{\arcsin{u}} \frac{1}{\sqrt{1 - u^2}} du$$