Question

For the following problems, use the substitutions \(\tan \left(\frac{x}{2}\right)=t, d x=\frac{2}{1+t^{2}} d t, \sin x=\frac{2 t}{1+t^{2}}\), and \(\cos x=\frac{1-t^{2}}{1+t^{2}}\). Find the area under the curve \(y=\frac{1}{1+\sin x}\) between \(x=0\) and \(x=\pi\). (Assume the dimensions are in inches.)

Short answer

The area under the curve is 2 square inches.

Step-by-step solution

Substitute and Transform the Integral Limits

Begin by noting that we want to find the area under the curve of the function.The integral is:\[\int_{0}^{\pi} \frac{1}{1 + \sin x} \, dx\]Using the substitutions \( \sin x = \frac{2t}{1+t^2} \) and \( dx = \frac{2}{1+t^2} \, dt \), the integral becomes:\[\int_{t_0}^{t_{\pi}} \frac{1}{1 + \frac{2t}{1+t^2}} \cdot \frac{2}{1+t^2} \, dt\]We need to calculate new limits of integration for \( t \). When \( x = 0 \), \( t_0 = \tan(0) = 0 \) and when \( x = \pi \), \( t_{\pi} = \tan\left(\frac{\pi}{2}\right) \), but noting that tangent becomes undefined at \( \frac{\pi}{2} \), analyze limits closely. Since we are using the Weierstrass substitution, proceed safely around the transformed results.

Simplify the Expression Inside the Integral

Simplify the integrand:First simplify the inner part:\[ 1 + \sin x = 1 + \frac{2t}{1+t^2} = \frac{1+t^2 + 2t}{1+t^2} = \frac{(1+t)^2}{1+t^2} \]So, the expression becomes:\[\frac{1}{1+\sin x} = \frac{1+t^2}{(1+t)^2}\]Now substitute back into the integral:\[\int_{0}^{\infty} \frac{1+t^2}{(1+t)^2} \cdot \frac{2}{1+t^2} \, dt = \int_{0}^{\infty} \frac{2}{(1+t)^2} \, dt\]

Evaluate the Simplified Integral

Evaluate the integral:\[ \int_{0}^{\infty} \frac{2}{(1+t)^2} \, dt \]This integral can be evaluated by recognizing it as a common form:Integrating, we have:\[-\frac{2}{1+t} \Bigg|_{0}^{\infty} \]Calculate the definite integral:- As \( t \to \infty, -\frac{2}{1+t} \to 0 \).- At \( t = 0, -\frac{2}{1+0} = -2 \).Thus, the area is:\[-(-2) = 2\]

Conclusion and Final Result

The evaluated integral gives the measure of the area under the curve.Hence, the area under the curve \( y = \frac{1}{1 + \sin x} \) from \( x = 0 \) to \( x = \pi \) is 2 square inches.

Key concepts

Integration Techniques

Integration techniques are essential tools in calculus that allow us to evaluate integrals that are not immediately obvious. In this exercise, the problem involves the use of trigonometric substitution to simplify the process of integration. Trigonometric substitution is particularly useful when dealing with integrands that contain radicals or trigonometric functions.

The substitution used here is based on the Weierstrass substitution, where we let \( \tan \left(\frac{x}{2}\right)=t \). This transformation helps to convert the trigonometric functions into rational expressions of \( t \), making the integration process more straightforward. This method is advantageous because it transforms a potentially complex integrand into a form that can be more easily managed with algebra.

The original integral is:

• \[ \int_{0}^{\pi} \frac{1}{1 + \sin x} \, dx \]

By substituting \( \sin x \) and \( dx \) into the integral using the Weierstrass substitution, the integral is transformed into an expression involving rational functions, which are generally easier to integrate:

• \[ \int_{0}^{\infty} \frac{2}{(1+t)^2} \, dt \]

This approach illustrates the power of integration techniques by leveraging variable transformations to simplify complex problems.

Definite Integrals

The concept of definite integrals is a key part of calculus, as they allow us to calculate the area under a curve between two points on the x-axis. In this exercise, we're tasked with finding the area under the curve \( y=\frac{1}{1+\sin x} \) from \( x=0 \) to \( x=\pi \).

To solve this, the integral was reconstructed through a substitution that changes the limits of integration as well. Instead of directly integrating with respect to \( x \), we change the variable to \( t \) using the Weierstrass substitution. This substitution also requires adjusting the limits of integration:

• When \( x = 0 \), \( t = \tan(0) = 0 \).
• When \( x = \pi \), \( t \to \infty \) because \( \tan \left(\frac{\pi}{2}\right) \) becomes undefined.

These new limits reflect how \( t \) changes as \( x \) moves from 0 to \( \pi \), and thus transforms the integral into:

• \[ \int_{0}^{\infty} \frac{2}{(1+t)^2} \, dt \]

Definite integrals summarize the total accumulation or the area over the specified interval, which in this case is calculated to be exactly 2 square inches.

Trigonometric Identities

Trigonometric identities play a vital role in simplifying integrals that involve trigonometric functions. When solving the given problem, key identities were used as substitutions to change the problem into a solvable form.

The problem particularly employed the identities related to the Weierstrass substitution:\[ \tan \left(\frac{x}{2}\right) = t \]and resulted in the following transformation formulas:

• \( \sin x = \frac{2t}{1+t^2} \)
• \( \cos x = \frac{1-t^2}{1+t^2} \)
• \( dx = \frac{2}{1+t^2} \, dt \)

By substituting \( \sin x \) into the integrand, \( \frac{1}{1+\sin x} \) was rewritten as a function of \( t \):

• \( \frac{1}{1+\frac{2t}{1+t^2}} = \frac{1+t^2}{(1+t)^2} \)

This essentially depurates the trigonometric function into a rational algebraic expression in terms of \( t \). Such identities are highly beneficial to streamline the calculation of integrals by reducing them to a more manageable form.

Understanding and using these identities efficiently allows us to turn extremely complicated integrals into simpler forms, which can then be integrated using standard techniques.