Question

Perform the indicated integrations. $$ \int \frac{(6 t-1) \sin \sqrt{3 t^{2}-t-1}}{\sqrt{3 t^{2}-t-1}} d t $$

Short answer

-2\cos(\sqrt{3t^2-t-1}) + C

Step-by-step solution

Identify a Substitution

To simplify the integral, we look for a substitution. Notice that the expression inside the square root, \(3t^2 - t - 1\), seems complex. If we let \(u = \sqrt{3t^2 - t - 1}\), then \(u^2 = 3t^2 - t - 1\). Our goal is to express everything in terms of \(u\).

Differentiate and Solve for dt

To change variables we differentiate \(u^2 = 3t^2 - t - 1\) with respect to \(t\), obtaining \(2u \frac{du}{dt} = 6t - 1\). Solving for \(dt\), we have \(dt = \frac{2u \, du}{6t - 1}\).

Substitute into the Integral

Replace \(\sqrt{3t^2 - t - 1}\) with \(u\) in the original integral to get: \(\int \frac{(6t-1) \sin(u)}{u} \cdot \frac{2u \, du}{6t-1}\). The \(6t-1\) terms cancel, simplifying the integral to \(2 \int \sin(u) \, du\).

Integrate \(\sin(u)\)

The integral \(2\int \sin(u) \, du\) is straightforward. Integrate \(\sin(u)\) to get \(-\cos(u)\), thus the integral becomes \(-2\cos(u) + C\), where \(C\) is the constant of integration.

Resubstitute \(u\)

Substituting back \(u = \sqrt{3t^2 - t - 1}\), the final solution is \(-2\cos(\sqrt{3t^2 - t - 1}) + C\).

Key concepts

Substitution Method

The Substitution Method is an essential tool in integration, particularly useful for simplifying complex integrals. It involves replacing a part of the integral with a new variable, effectively changing the variable's perspective. In the given exercise, the substitution method targets the expression inside the square root: \(3t^2 - t - 1\). The idea is to let this entire expression be \(u^2\), hence \(u = \sqrt{3t^2 - t - 1}\).

• Identify a part of the integrand that can be replaced to simplify the integral. Here, it's the expression inside the square root.
• Define a new variable \(u\) and substitute it. This new variable should simplify the calculation.

After making the substitution, differentiate \(u\) to express \(dt\) in terms of \(du\): that leads to \(dt = \frac{2u \, du}{6t - 1}\). This step is crucial as it reformulates the problem into a more manageable equation by eliminating the original variable \(t\), making the integral easier to solve.

Trigonometric Integration

Trigonometric Integration often occurs when dealing with integrals involving trigonometric functions like sine or cosine. In this case, after simplifying the integral using substitution, we are left with \(2 \int \sin(u) \, du\). This is a classic trigonometric integration problem, allowing us to directly integrate using standard rules.

• Recall that integrating \(\sin(u)\) results in \(-\cos(u)\).
• Always consider the constant of integration \(+ C\) for indefinite integrals.

The result here is quite straightforward: the integral \(2 \int \sin(u) \, du\) converts to \(-2\cos(u) + C\). The simplicity of this step emphasizes the power of substitution to transform a complex integral into an easily recognizable form.

Definite and Indefinite Integrals

The exercise concerns an indefinite integral, identifiable by its lack of upper and lower bounds. Indefinite integrals represent a family of functions and are pivotal as they include a constant of integration \(C\). This constant is necessary since differentiation of a function circumvents any additive constants, rendering the original integral's form unknown post-integration.

• Indefinite integrals provide general solutions that apply to a broad set of initial conditions.
• In contrast, definite integrals involve specific upper and lower limits, giving a numerical result instead.

When solving indefinite integrals, we're essentially determining an antiderivative. After substituting and integrating, resubstituting (as in the final step: substituting \(u\) back to \(\sqrt{3t^2 - t - 1}\)) is crucial for reverting to the original variable. This solution serves as the antiderivative for any given initial condition described in the problem context.