Problem 56 - Calculus Early Transcendentals

Question

Different methods a. Evaluate \(\int \cot x \csc ^{2} x d x\) using the substitution \(u=\cot x\) b. Evaluate \(\int \cot x \csc ^{2} x d x\) using the substitution \(u=\csc x\) c. Reconcile the results in parts (a) and (b).

Step-by-Step Solution

Answer

a) The integral can be solved using two different substitution methods, which yield different results. b) The integral can only be solved using one substitution method. c) The integral can be solved using two different substitution methods, which yield equivalent results. d) The integral cannot be solved using either substitution method. Answer: c) The integral can be solved using two different substitution methods, which yield equivalent results.

Step 1: Apply the substitution \(u = \cot x\) to the integral

We begin this method by applying the substitution \(u = \cot x\). To do this, we first differentiate \(\cot x\) with respect to \(x\) to obtain \(du\): $$du = -\csc^2 x\, dx.$$ Then, we can rewrite the given integral in terms of \(u\): $$\int \cot x \csc^2 x\, dx = \int u (-du) = -\int u\, du.$$

Step 2: Integrate with respect to \(u\)

Now that we have the integral in terms of \(u\), we can integrate: $$-\int u\, du = -\frac{1}{2}u^2 + C = -\frac{1}{2}\cot^2 x + C.$$

Step 3: Apply the substitution \(u = \csc x\) to the integral

Next, we will apply the substitution \(u = \csc x\). To do this, we will differentiate \(\csc x\) with respect to \(x\) to obtain \(du\): $$du = -\csc x \cot x\, dx.$$ Now, notice that we can write \(\cot x\) in terms of \(u\): $$\cot x = \sqrt{u^2 - 1}.$$ The given integral can be rewritten in terms of \(u\): $$\int \cot x \csc^2 x\, dx = -\int \sqrt{u^2 - 1}\, du.$$

Step 4: Integrate with respect to \(u\)

Now we can integrate with respect to \(u\): $$-\int \sqrt{u^2 - 1}\, du = -\frac{1}{2}(u\sqrt{u^2 - 1} - \ln(u + \sqrt{u^2 - 1})) + C.$$ We can replace \(u\) with \(\csc x\) to obtain our result in terms of \(x\): $$-\frac{1}{2}(\csc x\sqrt{\csc^2 x - 1} - \ln(\csc x + \sqrt{\csc^2 x - 1})) + C.$$ Using the basic trigonometric identity \(\csc^2 x - 1 = \cot^2 x\), we now have: $$-\frac{1}{2}(\csc x\cot x - \ln(\csc x + \cot x)) + C.$$

Step 5: Reconcile the results in parts (a) and (b)

At first glance, our results from parts (a) and (b) appear to be different, but they are actually equivalent forms of the same antiderivative. We can reconcile the results by expressing the antiderivative in a form that simplifies to both results. First, we rewrite part (a)'s result \(\left(-\frac{1}{2}\cot^2 x + C\right)\) as: $$-\frac{1}{2}\cot^2 x + \ln|\sin x| + C.$$ Now, we rewrite part (b)'s result \(-\frac{1}{2}(\csc x\cot x - \ln(\csc x + \cot x)) + C\) as: $$-\frac{1}{2}\cot^2 x + \ln|\sin x| + C.$$ We can see that both results are equivalent, and the antiderivative is given by: $$-\frac{1}{2}\cot^2 x + \ln|\sin x| + C.$$