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Chapter 5: Problem 34

Evaluate the integrals in Exercises \(13-48\) . $$ \int \csc \left(\frac{v-\pi}{2}\right) \cot \left(\frac{v-\pi}{2}\right) d v $$

Short Answer

Expert verified
The integral evaluates to \( -2 \csc \left( \frac{v-\pi}{2} \right) + C \).

Step by step solution

01

Recognize the Integral Form

We recognize that the integral \( \int \csc(x) \cot(x) \, dx \) is a standard form that has a direct antiderivative, \( -\csc(x) + C \). We will use this to help solve our integral.
02

Make a Substitution

Let \( u = \frac{v-\pi}{2} \), so \( du = \frac{1}{2} \, dv \), meaning that \( dv = 2 \, du \). We then rewrite the integral in terms of \( u \): \[ \int \csc(u) \cot(u) (2 \, du) \] which simplifies to \[ 2 \int \csc(u) \cot(u) \, du \].
03

Integrate using the Antiderivative

Using the known antiderivative from Step 1, we integrate \( 2 \int \csc(u) \cot(u) \, du \), which gives \[ -2 \csc(u) + C \].
04

Substitute Back to Original Variable

Substitute back \( u = \frac{v-\pi}{2} \): \[ -2 \csc \left( \frac{v-\pi}{2} \right) + C \].
05

Final Result

Therefore, the evaluated integral is \( -2 \csc \left( \frac{v-\pi}{2} \right) + C \), where \( C \) is the constant of integration.

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Key Concepts

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

Substitution Method
The substitution method is a technique used in calculus to simplify integration. It's like solving a puzzle by replacing complex parts with something simpler. If you see an integral that looks difficult, substitution can make it easier by changing the variable.

To perform a substitution, follow these steps:
  • Identify a part of the integral that can be substituted. Usually, this part will be within a function like sine, cosine, or tangent.
  • Find a suitable expression for substitution, often denoted by a new variable (usually "u").
  • Differentiate this expression to find "du" in terms of the original variable's differential.
  • Rewrite the integral using the substitution, which should simplify the integral into a more recognizable or solvable form.
In the example, we use substitution to simplify the integral of \( ext{csc}\left(\frac{v-\pi}{2}\right) \text{cot}\left(\frac{v-\pi}{2}\right) \) by setting \( u = \frac{v-\pi}{2} \). This reduces the complexity, making it easier to integrate.
Antiderivatives
Antiderivatives, also known as indefinite integrals, are functions that reverse the process of differentiation. Finding an antiderivative means finding a function whose derivative is the given function of the integral.

An antiderivative is always written with a constant of integration "C", because differentiation of a constant is zero and does not appear in the process of differentiation.
  • The general idea is to find a function that "undoes" the differentiation.
  • Each function can have multiple antiderivatives that differ by a constant.
  • The process of finding an antiderivative involves reversing the rules of differentiation.
In our specific example, the antiderivative of \( ext{csc}(u) ext{cot}(u) \) was identified as \( -\text{csc}(u) + C \). This step is crucial and simplifies the integration process once the correct form is used.
Csc and Cot Integrals
Integrating functions involving cosecant (\( \text{csc} \)) and cotangent (\( \text{cot} \)) might feel intimidating at first, given their less frequent use compared to sine and cosine. However, these functions have well-defined integral forms that can be tackled effectively.

The integral of \( \text{csc}(x) \text{cot}(x) \) is a standard form and has the direct antiderivative \( -\text{csc}(x) + C \). This means whenever you encounter such an integral, you can directly use the antiderivative.
  • Recognizing these integral forms helps solve problems quicker.
  • The values of \( \text{csc} \) and \( \text{cot} \) relate to the basic trigonometric function of sine and cosine.
  • Rely on these standard forms and substitutions when dealing with tricky integrals.
Once substitution has simplified an integral involving \( \text{csc} \) and \( \text{cot} \), as in our exercise, the known antiderivative can be applied to find the solution easily.

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Most popular questions from this chapter

Use the Substitution Formula in Theorem 6 to evaluate the integrals in Exercises \(1-24 .\) $$ \text { a. }\int_{0}^{1} r \sqrt{1-r^{2}} d r \quad \text { b. } \int_{-1}^{1} r \sqrt{1-r^{2}} d r $$

Suppose that $$\int_{0}^{1} f(x) d x=3$$ Find $$\int_{-1}^{0} f(x) d x$$ if a. \(f\) is odd, \(\quad \mathbf{b} . f\) is even.

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Suppose that \(F(x)\) is an antiderivative of \(f(x)=(\sin x) / x, x>0\) Express $$\int_{1}^{3} \frac{\sin 2 x}{x} d x$$ in terms of \(F .\)
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