Problem 55 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Solve the integral \(\int \tan x \sec ^{2} x dx\) using substitution methods with \(u = \tan x\) and \(u = \sec x\). Compare and reconcile the results. Answer: Both methods yield equivalent expressions for the antiderivative. With substitution \(u = \tan x\), the integral becomes \(\frac{\tan^2x}{2} + C\), and with substitution \(u = \sec x\), it becomes \(\frac{\sec^2x}{2} + C\). These expressions can be reconciled using the identity \(\tan^2 x = \sec^2 x - 1\) to show that they represent the same antiderivative.

Step 1: a. Integration with substitution \(u = \tan x\)

Let \(u = \tan x\). Differentiate both sides concerning \(x\): \(du = (\sec^2 x) dx\). Now we can replace \(\tan x\) by \(u\) and \(\sec^2 x dx\) by \(du\): $$\int u du$$ Now, integrate with respect to \(u\): $$\frac{u^2}{2}+ C = \frac{\tan^2x}{2} + C$$ The result of the first integral is \(\frac{\tan^2x}{2} + C\).

Step 2: b. Integration with substitution \(u = \sec x\)

Let \(u = \sec x\). We'll first rewrite the integral in terms of \(u\). Recall the identity: \(\tan^2 x = \sec^2 x - 1\). Then, \(\tan x = \sqrt{\sec^2 x - 1}\), so the integral becomes: $$\int \sqrt{u^2 - 1} \cdot u^2 du$$ Now, differentiate both sides concerning \(x\): \(du = (\sec x \tan x) dx\). Let's write \(\tan x\) in terms of \(u\): \(\tan x = \sqrt{u^2 - 1}\). Now our integral is: $$\int \frac{u^2\sqrt{u^2-1}}{u\sqrt{u^2-1}} du = \int u du$$ The integral is the same as in the part (a). Therefore, integrate with respect to \(u\): $$\frac{u^2}{2}+ C = \frac{\sec^2x}{2} + C$$ The result of the second integral is \(\frac{\sec^2x}{2} + C\).

Step 3: c. Reconciling the results

We obtained two different expressions for the integral of \(\tan x \sec^2 x dx\). The first expression is \(\frac{\tan^2x}{2} + C\), and the second expression is \(\frac{\sec^2x}{2}+ C\). Using the identity \(\tan^2 x = \sec^2 x - 1\), we can rewrite the first expression as: \(\frac{\sec^2 x -1}{2}+C\) Notice that the only difference between the two expressions is that the second one has an extra constant term 1/2, which can be absorbed into the constant of integration '\(C\)'. Therefore, these two expressions are equivalent and represent the same antiderivative.