Problem 46 - Calculus Early Transcendentals

Question

Integrals of cot \(x\) and \(\csc x\) Use a change of variables to prove that \(\int \cot x d x=\ln |\sin x|+C\).

Step-by-Step Solution

Answer

Answer: The integral of cot x is ln|sin x| + C.

Step 1: Define the change of variables

We will use u-substitution, so we will let \(u = \sin x\). Now, we need the derivative of \(u\) with respect to \(x\), which is \(\frac{du}{dx} = \cos x\).

Step 2: Find the differential for 'x'

We want to change dx to the du form. From step 1, we got \(\frac{du}{dx} = \cos x\). We can now find \(dx\) by dividing by the \(\cos x\): \(dx = \frac{du}{\cos x}\).

Step 3: Perform the change of variables

Now that we have \(u\) and \(dx\), we can substitute them into the integral. It becomes \(\int\frac{\cos x}{u} \cdot \frac{du}{\cos x}\). Notice that the \(\cos x\) terms cancel, which is what we want to simplify our integral. Now we have \(\int\frac{1}{u} du\).

Step 4: Solve the integral

We can now solve the integral. The integral of \(\frac{1}{u}\) with respect to \(u\) is simply \(\ln |u|+C\). So we have \(\ln |u|+C\).

Step 5: Reverse the change of variables

Now that we have the result in terms of \(u\), we need to reverse the change of variables to get our final answer in terms of \(x\). From step 1, we let \(u = \sin x\). So now, we have \(\ln |\sin x|+C\). The final answer is therefore \(\int \cot x dx=\ln |\sin x|+C\).

Key Concepts

U-Substitution

Understanding how u-substitution works is critical to solving many types of integral problems. Imagine it as the counterpart of the chain rule for differentiation, but in reverse. In the example of integrating cotangent, we define a new variable, typically labeled as 'u', that simplifies the original integral.

The choice of what 'u' should be is often guided by what makes the integral simpler. In the case of \( \cot x \), we let \( u = \sin x \). Once 'u' is defined, we then calculate the derivative of 'u' with respect to 'x' to establish a relationship between \( dx \) and \( du \) which helps us to convert the original integral into one in terms of 'u'. This makes the integral more straightforward and often allows direct application of integration rules.

However, choosing 'u' requires practice and sometimes a bit of insight, as it can make the difference between a solvable problem and one that's not immediately clear.

Integration Techniques

Once we are familiar with u-substitution, we can delve into other integration techniques that are used to tackle more complex functions. These techniques include, but are not limited to, integration by parts, partial fraction decomposition, and trigonometric integration.

Each technique has its own specific situations where it is the most effective. For instance, integration by parts is particularly useful when dealing with the product of two functions, while partial fraction decomposition is ideal for complex rational functions.

The integral of \( \cot x \) simplifies nicely with u-substitution, but knowing other techniques broadens the range of integrals one can solve. Mastery of these methods not only improves problem-solving skills but also deepens understanding of the intricacies of calculus.

Inverse Trigonometric Functions

When dealing with inverse trigonometric functions, we're often faced with integrating functions that are the derivatives of these inverses. For example, the integral of \( \cot x \) leads us to the natural logarithm of the absolute value of sine, which can be thought of as an inverse operation to differentiation of an inverse trigonometric function.

It's important to recognize that the natural logarithm functions appear frequently when integrating trigonometric functions due to their properties and relationship with exponentiation and the exponential function. These connections are part of why trigonometry and calculus are so deeply intertwined and serve to broaden the toolkit of anyone delving into integral calculus.

Definite Integrals

Moving beyond finding antiderivatives, definite integrals allow us to calculate the actual 'area under the curve' for a specified interval. Unlike indefinite integrals which provide the general form of the antiderivative plus a constant of integration, \( C \), definite integrals are numbers representing the total accumulation over the given interval.

Definite integration involves evaluating the integral from a lower bound 'a' to an upper bound 'b' and has powerful applications in physics, engineering, and economics. When solving for a definite integral, the process remains largely the same as for an indefinite integral. However, after finding the antiderivative, we find the difference in its values at 'b' and 'a', often referred to as 'evaluating the bounds'. This subtracts out the constant of integration and yields the net change or total accumulation over the interval.