Problem 41 - 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 function $$2\sec^2(2v)$$ with respect to \(v\) and verify the result by differentiation. Solution: The indefinite integral of the given function is $$\tan(2v) + C$$. This has been verified by differentiation, as the derivative of our result matches the original function.

Step 1: Integrate the given function with respect to v

To compute the indefinite integral, we can use the following formula for integrating trigonometric functions: $$\int \sec^2(ax)dx = \frac{1}{a} \tan (ax) + C$$ Here, $$a=2$$, so the function becomes: $$\int 2\sec^2(2v)dv = 2\int \sec^2(2v)dv$$ Applying the formula above, we have: $$2\int \sec^2(2v)dv = 2\left(\frac{1}{2}\tan(2v)\right) + C$$ Simplifying, we get: $$\int 2\sec^2(2v)dv = \tan(2v) + C$$

Step 2: Verify the result by differentiation

Now, we will take the derivative of our result to verify whether it matches the original function or not. We have: $$\frac{d}{dv}(\tan(2v) + C)$$ Using the chain rule, we get: $$\frac{d}{dv}(\tan(2v)) = \sec^2(2v) \cdot \frac{d}{dv}(2v)$$ Which simplifies to: $$\sec^2(2v) \cdot 2 = 2\sec^2(2v)$$ So, the derivative of our result matches the original function, which confirms that our solution for the indefinite integral is correct.

Key Concepts

Trigonometric Functions

Trigonometric functions are fundamental in calculus, often appearing in integration and differentiation problems. They describe relationships in triangles and include functions like sine (\(\sin\)), cosine (\(\cos\)), and tangent (\(\tan\)). In this exercise, the trigonometric function used is the secant squared (\(\sec^2\)). Understanding the role of these functions in calculus is crucial because they commonly appear in problems involving waves, oscillations, and many physical phenomena.

The secant function is the reciprocal of the cosine function, given as:

• \(\sec(x) = \frac{1}{\cos(x)}\)

When squared, \(\sec^2(x)\) becomes particularly important because its integral leads us directly to the tangent function, which is seen in the formula used in this problem:

• \(\int \sec^2(ax)dx = \frac{1}{a} \tan (ax) + C\)

Knowing these relationships allows us to integrate naturally occurring trigonometric expressions with ease.

Integration by Substitution

Integration by substitution is a popular method in calculus for simplifying integrals. It's akin to the reverse of the chain rule used in differentiation. In this exercise, the substitution isn't explicitly shown, but it's implied through the application of the integration formula.

We have the integral \(\int 2 \sec^2(2v) dv\). Here, the substitution approach would involve setting a new variable, such as \(u = 2v\), leading to \(du = 2 \cdot dv\). This substitution simplifies the computation:

• Rewriting the integral: \(\int \sec^2(u) du\)

Using the formula \(\int \sec^2(u) du = \tan(u) + C\), we solve for the integral in terms of \(u\). You then convert back to the original variable to finish the integration. This method helps tackle complex integrals by converting them into more manageable forms.

Integration Formula

Integration formulas are indispensable tools in solving indefinite integrals. They provide a way to directly solve integrals by expressing them in terms of known functions. The formula applied in the step-by-step solution here was:

\[\int \sec^2(ax)dx = \frac{1}{a} \tan (ax) + C\] This formula is derived from the differentiation of the tangent function. When differentiating \(\tan(ax)\), the result is \(\sec^2(ax) \cdot a\). Hence, integrating \(\sec^2(ax)\) returns us to the tangent function, scaled by \(\frac{1}{a}\).

• The constant \(C\) signifies the arbitrary constant of integration, which accounts for any constant that could be added during differentiation.

By understanding and memorizing these formulas, solving trigonometric integrals becomes a streamlined process, allowing students to quickly find solutions that might otherwise require lengthy calculations.