Problem 38 - Calculus of a Single Variable

Question

Finding an Indefinite Integral of a Trigonometric Function In Exercises \(31-40\) , find the indefinite integral. $$ \int \frac{\csc ^{2} t}{\cot t} d t $$

Step-by-Step Solution

Answer

The indefinite integral of \(\int \frac{\csc ^{2} t}{\cot t} d t\) is \(-ln|cot(t)|+C\)

Step 1: Simplify the given integral

Simplify the integral by observing the properties of cotangent(\(cot(t)\)) and cosecant (\(csc(t)\)). We know that the \(cot(t)= \frac{1}{tan(t)}\) and \(csc(t) = \frac{1}{sin(t)}\). The integral can be written as: \(\int \frac{csc^2(t)}{cot(t)} dt = \int \frac{1}{\(sin^2(t)\)} \frac{1}{\frac{1}{tan(t)}} dt = \int \frac{tan(t)}{sin^2(t)} dt = \int \frac{\frac{sin(t)}{cos(t)}}{sin^2(t)} dt = \int \frac{1}{sin(t) cos(t)} dt)\)

Step 2: Use substitution method

Now, identify a function and its derivative in the integrand such that the integrand can be written in the simpler form u' du , for some function u. Notice that the derivative of \(cot(t)\) is \(-csc^2(t)\). So, let's set \(u=cot(t)\) . Then the differential \(du = -csc^2(t) dt\). So \(dt = - du / csc^2(t)\) = - du / u'\)

Step 3: Substitution into the Integral

Substitute \(u = cot(t)\) and \(du = - csc^2(t) dt\) into the integral to get\[ \int - \frac{1}{u sin(t) cos(t)} du = - \int \frac{1}{u} du \]

Step 4: Finding the Resulting Integral

The integral of \(1/u\) is \(ln|u|\), therefore: \[ -\int \frac{1}{u} du = -ln|u|+C \]

Step 5: Conversion back to Original Variable

Convert back to original variable \(t\). Remember that we have set \(u=cot(t)\), so the final answer is: \[ -ln|cot(t)|+C \]

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used to simplify integrals that contain square roots and trigonometric functions. By using trigonometric identities, we express complex algebraic expressions in terms of simpler trigonometric functions.

For instance, if we have an expression involving \(\sqrt{1-x^2}\), we can substitute \(x = \sin(\theta)\), taking advantage of the Pythagorean identity \(1-\sin^2(\theta)=\cos^2(\theta)\). Similarly, in our exercise, the presence of cosecant and cotangent suggests we can simplify the integral with appropriate trigonometric substitutions.

Integration by Substitution

Integration by substitution, also known as u-substitution, is akin to the reverse process of the chain rule in differentiation.

It involves choosing a part of the integrand to be \(u\), then finding \(du\) and substituting in to make the integral simpler. This works well when you can identify a function and its derivative within the integral. In our example, identifying \(u = \cot(t)\) and \(du = -\csc^2(t) dt\) streamlines the integral into a form that can be easily integrated.

Indefinite Integral

An indefinite integral, often referred to as an antiderivative, represents a family of functions whose derivative is the integrand. In simple terms, integrating a function is the reverse process of differentiating it.

The notation \(\int f(x) dx\) implies finding all functions \(F(x)\) such that \(F'(x) = f(x)\). Indefinite integrals come along with an arbitrary constant \(C\), symbolizing that any number can be added to the antiderivative and still represent the original function. For instance, the indefinite integral of our trigonometric function will include a constant of integration, \(C\), in the final result.

Cosecant Squared

The function \(\csc^2(t)\), or cosecant squared, comes from \(\csc(t)\) which is the reciprocal of the sine function.

Specifically, \(\csc(t) = \frac{1}{\sin(t)}\), and thus \(\csc^2(t) = \frac{1}{\sin^2(t)}\). This trigonometric function is useful in integration due to its relationship with the derivative of \(\cot(t)\). This integral is useful in many areas of calculus and appears often in problems involving trigonometric identities.

Cotangent

Cotangent, denoted as \(\cot(t)\), is a trigonometric function representing the reciprocal of the tangent function. It is defined as \(\cot(t) = \frac{\cos(t)}{\sin(t)}\), or alternatively, \(\cot(t) = \frac{1}{\tan(t)}\).

In integration, \(\cot(t)\) is particularly useful when its derivative, which is \(\-\csc^2(t)\), appears in an integral; this is the key insight in solving the given exercise. By recognizing this, we can implement integration by substitution effectively, which significantly simplifies the integral and brings us closer to finding the final solution.