Problem 40 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Find the indefinite integral of the given function and verify the result: $$\int 2 \sec ^{2} 2 v d v$$ Solution: The indefinite integral of the given function is: $$\int 2 \sec ^{2} 2 v d v = \tan(2v) + C$$ Verification: The derivative of the antiderivative, $$\frac{d}{dv} (\tan(2v) + C)$$, gives the original function $$2\sec^2(2v)$$.

Step 1: Recognize the integral

We are given the integral: $$\int 2 \sec ^{2} 2 v d v$$ Our goal is to find an antiderivative, which when differentiated, returns the original function.

Step 2: Perform the integration

The function is already prepared for direct integration, so we can just integrate with respect to \(v\): $$\int 2 \sec ^{2} 2 v d v$$ Recall that the derivative of the tangent function is the secant squared function: $$\frac{d}{dx} (\tan(x)) = \sec^2(x)$$ In our case, we have \(\sec^2(2v)\). We can use substitution to find the integral. Let \(u = 2v\), then \(\frac{du}{2} = dv\). Our integral then becomes: $$\int 2 \sec^2(u) \frac{du}{2}$$ This simplifies to: $$\int \sec^2(u) du$$ Now, this is a standard integral and integrates to: $$\tan(u) + C$$ Now, we need to substitute back and get our answer in terms of \(v\): $$\tan(2v) + C$$ So, the indefinite integral of the given function is: $$\int 2 \sec ^{2} 2 v d v = \tan(2v) + C$$

Step 3: Check the result by differentiation

Now, let's check our work by differentiating the antiderivative: $$\frac{d}{dv} (\tan(2v) + C)$$ We use the chain rule and the fact that the derivative of tangent is secant squared: $$\frac{d}{dv} (\tan(2v)) = 2\sec^2(2v)$$ As expected, our antiderivative correctly differentiates back to the original function: $$2\sec^2(2v)$$ The solution is verified.