Question

Evaluate the following integrals as they are written. $$\int_{0}^{\pi / 2} \int_{0}^{\cos y} e^{\sin y} d x d y$$

Short answer

Question: Evaluate the double integral: $$\int_{0}^{\pi / 2} \int_{0}^{\cos y} e^{\sin y} d x d y$$ Solution: After evaluating the double integral step by step, we find that the value of the given double integral is $$\ln(1 + \sqrt{2})$$.

Step-by-step solution

Evaluate the inside integral with respect to x

Let's first evaluate the inner integral with respect to x: $$\int_0^{\cos y} e^{\sin y} dx$$ In this integral, \(e^{\sin y}\) is a constant with respect to x, so we can take it out of the integral: $$e^{\sin y} \int_0^{\cos y} dx$$ Now, we integrate with respect to x: $$e^{\sin y}(x\Big|_0^{\cos y}) = e^{\sin y} [\cos y- 0] = e^{\sin y}\cos y$$

Evaluate the outside integral with respect to y

Now we will evaluate the outside integral with respect to y: $$\int_0^{\pi/2} e^{\sin y}\cos y dy$$ To evaluate this integral, we will use integration by substitution. Let: $$u = \sin y$$ Then, $$\frac{du}{dy} = \cos y$$ $$dy = \frac{du}{\cos y}$$ By this substitution, our limits will also change. When \(y = 0\), \(u = \sin 0 = 0\), and when \(y = \pi / 2\), \(u = \sin(\pi/2) = 1\). So, the new integral is: $$\int_0^1 e^u \frac{du}{\cos y(u)}$$ \(\cos y(u)\) can be expressed using u since \(u = \sin y\), and we can use the Pythagorean identity: $$\cos y(u) = \sqrt{1 - u^2}$$ Now, the integral becomes: $$\int_0^1 e^u \frac{du}{\sqrt{1 - u^2}}$$ It becomes a standard integral (sinh-1 integral): $$\int_0^1 \frac{e^u}{\sqrt{1 - u^2}}du$$ You can look up its value in a table of standard integrals, which gives the result: $$\sinh^{-1}(1) = \ln(1 + \sqrt{2})$$ So, the value of the given double integral is: $$\int_{0}^{\pi / 2} \int_{0}^{\cos y} e^{\sin y} d x d y = \ln(1 + \sqrt{2})$$

Key concepts

Integration by Substitution

Understanding integration by substitution, also known as u-substitution, is essential for solving complex integrals. This method involves changing the variable of integration to simplify the integral. It’s akin to substituting players in a sports game to better tackle the situation on the field.

Imagine an integral as a complicated puzzle; u-substitution lets you transform it into a more familiar shape that's easier to solve. You choose a part of the integral to be your new variable, u, and find the derivative, which gives you du. This allows you to replace parts of the integral with u and du, simplifying your integral in the process.

However, one must be careful: the limits of integration will also change if you’re dealing with a definite integral. Make sure to calculate the new limits based on your substitution. In our exercise, \( u = \sin y \) makes it easier to work with the integral because the new variable u directly involves the function within the exponential, and its derivative \( \frac{du}{dy} = \cos y \) is already present in the integral.

Definite Integrals

A definite integral has upper and lower limits and calculates the net area under a curve. Picture it as finding the total space a fence will enclose within certain boundaries in your yard. You're not just looking at the shape of the fence but actually measuring the enclosed area.

The integral from the exercise \( \int_{0}^{\pi / 2} \int_{0}^{\cos y} e^{\sin y} d x d y \) is a type of definite integral. The outer integral has limits from 0 to \( \pi / 2 \) and the inner one from 0 to \( \cos y \). When evaluating definite integrals, after finding the antiderivative, you subtract the value of the antiderivative at the lower limit from the value at the upper limit to find that enclosed area.

Pythagorean Identity

The Pythagorean identity is a cornerstone of trigonometry, derived from the Pythagorean theorem you’d find in a right triangle. It states that for any angle \( y \), \( \sin^2 y + \cos^2 y = 1 \). It’s an incredibly powerful tool for transforming trigonometric expressions, especially in integrals involving sine and cosine.

In the double integral exercise, after u-substitution, we needed to transform \( \cos y \) in terms of u, and the Pythagorean identity came to the rescue. By using \( u = \sin y \) and the identity, we found that \( \cos y(u) = \sqrt{1 - u^2} \). This transformation is vital as it changes the integral into a form that matches a standard integral—a form we can solve using known techniques or reference materials.

Standard Integrals

Standard integrals are the bread and butter of integration. These integrals have been solved and tabulated because they appear frequently. They serve as a reference to quickly solve parts of more complex integrals once they have been transformed into a recognizable form. It's like having a cheat sheet of solutions for specific types of puzzles.

In the provided solution, we arrive at a point where the integral \( \int_0^1 \frac{e^u}{\sqrt{1 - u^2}}du \) matches a standard integral form involving the inverse hyperbolic sine function. These standard forms are conveniently listed in tables, so we don't have to integrate them from scratch. In this case, we use the table to find that this standard integral evaluates to \( \sinh^{-1}(1) = \ln(1 + \sqrt{2}) \) which is the final answer for our double integral. Knowing your standard integrals can save you a lot of time and effort for certain types of problems.