91Ó°ÊÓ

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.

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: $$\int 2 \sec ^{2} 2 v d v = \tan(2v) + C$$

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.