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Find the indefinite integral. $$ \int \frac{\cos t}{1+\sin t} d t $$

Short Answer

Expert verified
\( \ln |1+\sin t| + C \)

Step by step solution

01

Identify a Suitable Substitution

One efficient way to simplify this integrand is through trigonometric identity. This requires choosing \( u = 1 + \sin t \). This is a useful substitution because the derivative of \( \sin t \) is \( \cos t \), which appears in the numerator of the integrand.
02

Substitute

Next, make the substitution into the integral. The derivative \( du = \cos t \, dt \) is also substituted. The integral becomes \[ \int \frac{1}{u} \, du \]
03

Solve the Simplified Integral

The integral \( \int \frac{1}{u} \, du \) is a well-known standard integral, whose antiderivative is \( \ln |u| + C \) where \( C \) represents the constant of integration. Upon substitution, we get \( \ln |1+\sin t| + C \)锔.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Substitution
Trigonometric substitution is a technique used in integration to simplify complex expressions involving square roots and certain trigonometric functions. In our exercise, \[ \int \frac{\cos t}{1+\sin t} dt \], the integrand is simplified by using trigonometric substitution, which involves replacing a part of the integrand with a trigonometric function to make integration manageable.

In this case, the substitution made is \( u = 1 + \sin t \), a strategic choice because the derivative of \( \sin t \) is \( \cos t \), matching the numerator. The goal is to transform the integral into one that is easier to handle. Often, trigonometric identities like \( \sin^2(t) + \cos^2(t) = 1 \) are used, but in this exercise, the substitution alone aligns with the derivative to simplify the integral.
Antiderivative
The antiderivative, or the indefinite integral, is essentially the reverse of differentiation. When speaking of antiderivatives, we refer to a function whose derivative gives the original function. For instance, if \( F'(x) = f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \). In our exercise, once the integral is simplified to \[ \int \frac{1}{u} du \], the antiderivative is readily recognized as the natural logarithm function. This is because the derivative of \( \ln|u| \) is \( 1/u \).

Remember, when finding an antiderivative, a constant of integration, denoted as \( C \), is added since the derivative of a constant is zero. It represents the family of all antiderivatives of the given function.
Integration Techniques
There are various integration techniques employed to solve integrals, ranging from direct integration of basic functions to more sophisticated methods for complex functions. These include substitution, integration by parts, partial fractions, and trigonometric substitution among others.

In the given exercise, substitution is the primary technique used. This involves changing the variable to simplify the integral into a more recognizable form. After substitution, we recognize the new integral as a basic form where direct integration is possible. Knowing certain standard integrals is crucial as they often appear after employing these techniques, enabling us to quickly identify and solve them. For example, the standard integral \[ \int \frac{1}{x} dx = \ln|x| + C \] is utilized in this exercise following our strategic substitution.

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Most popular questions from this chapter

Vertical Motion An object is dropped from a height of 400 feet. (a) Find the velocity of the object as a function of time (neglect air resistance on the object). (b) Use the result in part (a) to find the position function. (c) If the air resistance is proportional to the square of the velocity, then \(d v / d t=-32+k v^{2}\), where \(-32\) feet per second per second is the acceleration due to gravity and \(k\) is a constant. Show that the velocity \(v\) as a function of time is \(v(t)=-\sqrt{\frac{32}{k}} \tanh (\sqrt{32 k} t)\) by performing the following integration and simplifying the result. \(\int \frac{d v}{32-k v^{2}}=-\int d t\) (d) Use the result in part (c) to find \(\lim _{t \rightarrow \infty} v(t)\) and give its interpretation. (e) Integrate the velocity function in part (c) and find the position \(s\) of the object as a function of \(t\). Use a graphing utility to graph the position function when \(k=0.01\) and the position function in part (b) in the same viewing window. Estimate the additional time required for the object to reach ground level when air resistance is not neglected. (f) Give a written description of what you believe would happen if \(k\) were increased. Then test your assertion with a particular value of \(k\).

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