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Chapter 4: Problem 40

Determine the following indefinite integrals. Check your work by differentiation. $$\int 2 \sec ^{2} 2 v d v$$

Short Answer

Expert verified
Question: Evaluate the indefinite integral $$\int 2 \sec^2 2v dv$$ and check your work by differentiation. Answer: The indefinite integral is $$\int 2 \sec^2 2v dv = \frac{1}{2} (\tan (2v) + C)$$, and when differentiated, it gives back the original integrand $$2\sec^2(2v)$$.

Step by step solution

01

Perform substitution

Let \(u = 2v\). Then, \(du = 2dv\). Now, we can rewrite the integral in terms of \(u\): $$\int 2 \sec^2 2v dv = \int \sec^2 u \frac{1}{2} du$$
02

Solve the integral in terms of u

Now, we can evaluate the simpler integral in terms of \(u\): $$\int \sec^2 u \frac{1}{2} du = \frac{1}{2} \int \sec^2 u du$$ Recall that the integral of \(\sec^2 u\) is \(\tan u\) (plus an arbitrary constant \(C\)). Then, we have: $$\frac{1}{2} \int \sec^2 u du = \frac{1}{2} (\tan u + C)$$
03

Substitute back

Since we originally set \(u = 2v\), we now substitute back to find the integral in terms of \(v\): $$\frac{1}{2} (\tan u + C) = \frac{1}{2} (\tan (2v) + C)$$ Hence, the indefinite integral is: $$\int 2 \sec^2 2v dv = \frac{1}{2} (\tan (2v) + C)$$
04

Check by differentiation

Now, let's differentiate our result with respect to \(v\) to make sure we get back the original integrand: $$\frac{d}{dv} \left(\frac{1}{2} (\tan (2v) + C)\right)$$ Using the chain rule, we have: $$\frac{d}{dv} \left(\frac{1}{2} (\tan (2v) + C)\right) = \frac{1}{2} (2 \sec^2 (2v)) = 2\sec^2(2v)$$ Since the derivative of our result matches the original integrand, we have successfully evaluated the indefinite integral and checked our work.

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