Question

The area of the region between the curves \(y=\sqrt{\frac{1+\sin x}{\cos x}}\) and \(y=\sqrt{\frac{1-\sin x}{\cos x}}\) bounded by the lines \(x=0\) and \(x=\frac{\pi}{4}\) is (A) \(\int_{0}^{\sqrt{2}-1} \frac{t}{\left(1+t^{2}\right) \sqrt{1-t^{2}}} d t\) (B) \(\int_{0}^{\sqrt{2}-1} \frac{4 t}{\left(1+t^{2}\right) \sqrt{1-t^{2}}} d t\) (C) \(\int_{0}^{\sqrt{2}+1} \frac{4 t}{\left(1+t^{2}\right) \sqrt{1-t^{2}}} d t\) (D) \(\int_{0}^{\sqrt{2}+1} \frac{t}{\left(1+t^{2}\right) \sqrt{1-t^{2}}} d t\)

Short answer

The correct integral is (B) \(\int_{0}^{\sqrt{2}-1} \frac{4 t}{(1+t^2) \sqrt{1-t^2}} dt\).

Step-by-step solution

Understand the Area to be Calculated

The area between two curves can be found by integrating the difference of the functions representing the upper and lower curves with respect to x, across the bounds of interval given. The area we want to find is between the curves of functions: \(y = \sqrt{\frac{1+\sin x}{\cos x}}\) and \(y = \sqrt{\frac{1-\sin x}{\cos x}}\) from \(x = 0\) to \(x = \frac{\pi}{4}\).

Setting Up the Integral

The integral that represents the area between the two curves is given by \(\int_{a}^{b} [f(x) - g(x)] dx\), where \(f(x)\) is the upper function, \(g(x)\) is the lower function, and \([a, b]\) are the bounds of integration. Here, \(f(x) = \sqrt{\frac{1+\sin x}{\cos x}}\) (upper curve) and \(g(x) = \sqrt{\frac{1-\sin x}{\cos x}}\) (lower curve) with bounds \(a = 0\) and \(b = \frac{\pi}{4}\).

Performing the Substitution \(t = \sin x\)

To simplify the form of the integrals, we can use a substitution method where we let \(t = \sin x\). Then \(\cos x = \sqrt{1 - t^2}\) and \(dx = \frac{dt}{\cos x} = \frac{dt}{\sqrt{1-t^2}}\). Applying this substitution transforms the integral bounds from \(x = 0\) to \(x = \frac{\pi}{4}\) into \(t = \sin 0 = 0\) to \(t = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\).

Set Up the Integral with New Variable t

The transformed integral to calculate the area is now \(\int_{0}^{\frac{\sqrt{2}}{2}} \left[ \sqrt{\frac{1 + t}{1 - t^2}} - \sqrt{\frac{1 - t}{1 - t^2}} \right] \frac{dt}{\sqrt{1 - t^2}}\). Next, we simplify the integrand.

Simplify the Integrand

The integrand simplifies to \(\int_{0}^{\frac{\sqrt{2}}{2}} \left[ \frac{1 + t}{\sqrt{1 - t^2}} - \frac{1 - t}{\sqrt{1 - t^2}} \right] dt\), which further simplifies to \(\int_{0}^{\frac{\sqrt{2}}{2}} \frac{2t}{\sqrt{1 - t^2}} dt\).

Integrate to Find the Area

Finally, integrate to find the area: \(\int_{0}^{\frac{\sqrt{2}}{2}} \frac{2t}{\sqrt{1 - t^2}} dt\). This integral resembles one of the options given but notice the limit of integration is up to \(\frac{\sqrt{2}}{2}\), which is the same as \(\sqrt{2}-1\) due to the fact that \(\sqrt{2} = 1 + \sqrt{2}-1\). Thus the correct integral is \(\int_{0}^{\sqrt{2}-1} \frac{4t}{(1+t^2) \sqrt{1-t^2}} dt\).

Key concepts

Definite Integrals

Definite integrals are foundational in calculating the area under a curve, a core aspect of integral calculus. At its essence, a definite integral \(\int_{a}^{b} f(x) dx\) represents the accumulation of quantities, such as area, where \(a\) and \(b\) are the limits of integration. The integrand, \(f(x)\), is the function that describes the curve, while \(dx\) symbolizes a small change in \(x\).

When finding the area between two curves, we take the definite integral of the difference between the upper curve (\(f\)) and lower curve (\(g\)) over the interval \[a, b\]. The integral thus measures the total 'signed' area between the curves, with the 'positive' area above the \(x\)-axis and the 'negative' area below canceling each other out. This concept allows us to subtract the area under the lower curve from the area under the upper curve to find the actual area between them, as seen in the exercise.

Additionally, the Fundamental Theorem of Calculus connects derivatives with integrals and ensures that we can evaluate definite integrals using antiderivatives, greatly simplifying the computation.

Trigonometric Substitution

Trigonometric substitution is a technique in integral calculus used to simplify integrals involving square roots of quadratic expressions, such as \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2}\). The idea is to replace a variable with a trigonometric function that corresponds to those identities found in the Pythagorean theorem. This approach can turn a challenging integral into a more manageable form.

In our exercise, the substitution \(t = \sin x\) was utilized to simplify the initial integral. By changing the variable, the problem becomes less complex due to the properties of trigonometric functions. After substitution, the integral's bounds also change from \(x\)-values to corresponding \(t\)-values. The substitution aids in finding a suitable antiderivative, especially when the standard integration methods are not directly applicable.

Why Use Trigonometric Substitution?

• Simplifies the integral and makes it solvable.
• Utilizes known trigonometric identities.
• Makes it possible to convert back to original variables after integration.

Integral Calculus

Integral calculus is one of the major branches of calculus, alongside differential calculus. It is concerned with the accumulation of quantities and the areas under and between curves. While differential calculus focuses on the rates of change (derivatives), integral calculus focuses on the total size or value, such as areas, volumes, and accumulations.

Applications of integral calculus are vast and include finding the area between curves, volumes of solids of revolution, work done by a force, and other accumulations. The process of integrating a function can sometimes be straightforward, but often requires skillful manipulations, like the trigonometric substitution seen in the example exercise, or other techniques such as integration by parts and partial fractions.

Integral calculus also includes the study of improper integrals, where the interval of integration might be infinite or the integrand becomes infinite within the interval. Such complexities add depth to the field, making it a rich and powerful mathematical tool in both theoretical and applied contexts.