Problem 63 - Thomas Calculus

Question

Use reduction formulas to evaluate the integrals. \(\int 8 \cot ^{4} t d t\)

Step-by-Step Solution

Answer

\(-\frac{8}{3} \cot^3 t - 8 \cot t - 8t + C\)

Step 1: Identify the Reduction Formula

To solve the integral \( \int 8 \cot^4 t \, dt \) using reduction formulas, we need a formula that reduces the power of \( \cot t \). The common reduction formula used for the integration of \( \cot^n x \) is based on the recursion \[ \int \cot^n x \, dx = -\frac{1}{n-1} \cot^{n-1} x + \int \cot^{n-2} x \, dx \] when \( n \geq 2 \).

Step 2: Apply the Reduction Formula

Apply the reduction formula to the integral. For \( \int 8 \cot^4 t \, dt \), using the formula, we get:\[ \int \cot^4 t \, dt = -\frac{1}{4-1} \cot^3 t + \int \cot^2 t \, dt = -\frac{1}{3} \cot^3 t + \int \cot^2 t \, dt \]

Step 3: Simplify \( \int \cot^2 t \, dt \)

Now simplify \( \int \cot^2 t \, dt \). Since \( \cot^2 t = \csc^2 t - 1 \), we can write:\[ \int \cot^2 t \, dt = \int (\csc^2 t - 1) \, dt = \int \csc^2 t \, dt - \int 1 \, dt \]The integral \( \int \csc^2 t \, dt \) is \( -\cot t \) while \( \int 1 \, dt \) is \( t \). Thus,\[ \int \cot^2 t \, dt = -\cot t - t \]

Step 4: Combine Results

Substitute the result from Step 3 back into the equation from Step 2:\[ \int \cot^4 t \, dt = -\frac{1}{3} \cot^3 t - \cot t - t \]

Step 5: Multiply by the Constant

Don't forget to multiply the result by 8, since the original integral is \( \int 8 \cot^4 t \, dt \). Thus,\[ \int 8 \cot^4 t \, dt = 8 \left( -\frac{1}{3} \cot^3 t - \cot t - t \right) = -\frac{8}{3} \cot^3 t - 8 \cot t - 8t \]

Step 6: Add the Constant of Integration

Finally, add the constant of integration \( C \) to account for the indefinite integral:\[ \int 8 \cot^4 t \, dt = -\frac{8}{3} \cot^3 t - 8 \cot t - 8t + C \]

Key Concepts

Reduction Formulas

Reduction formulas in integral calculus are helpful tools for evaluating integrals of functions with powers. Specifically, they transform complex expressions into simpler ones by reducing the power of a function to aid in integration.
These formulas are particularly useful when dealing with trigonometric functions. Reduction formulas work on the principle of recursion. They allow you to express an integral of a function with higher powers in terms of an integral with lower powers.

• In the given exercise, we needed to integrate \( \int \cot^4 t \, dt \) using a reduction formula specific to the cotangent function.
• The typical reduction formula for \( \int \cot^n x \, dx \) is \[ \int \cot^n x \, dx = -\frac{1}{n-1} \cot^{n-1} x + \int \cot^{n-2} x \, dx \] for \( n \geq 2 \).

This recursion reduces the degree of cotangent, making the integral manageable and solvable. Understanding and applying reduction formulas efficiently can significantly simplify the problem of integrating complex trigonometric functions.

Trigonometric Integrals

Trigonometric integrals involve integrating functions that include trigonometric terms. These integrals are common in mathematical problems and often appear in the form of powers or products of trigonometric functions.
This category requires special techniques and strategies to solve effectively due to the oscillatory nature of trigonometric functions.

• In the given problem, we dealt with the integral of \( \cot^4 t \), a trigonometric function raised to a power.
• We reduced the integral using the reduction formula, simplifying complex trigonometric powers into more straightforward components like \( \cot^3 t \) and \( \cot^2 t \).

When solving trigonometric integrals, it’s essential to recognize identities such as \( \cot^2 t = \csc^2 t - 1 \), which help break down the integral. Such identities are fundamental in rewriting expressions and finding integrals in closed forms. Mastering these concepts allows students to tackle a broad range of problems confidently.

Indefinite Integrals

Indefinite integrals represent a family of functions whose derivative is the given integrand. Unlike definite integrals, indefinite integrals do not have upper and lower limits. Instead, they include a constant of integration \( C \) to cover all possible antiderivatives.
They are crucial for solving problems where finding the original function is needed from its derivative.

• In the problem, after calculating \( \int 8 \cot^4 t \, dt \), we obtained \[ -\frac{8}{3} \cot^3 t - 8 \cot t - 8t + C \]
• Here, \( C \) is added at the end, indicating the indefinite nature of the integral.

Recognizing the indefinite integral as a means to generalize all possible solutions is fundamental in calculus. This practice reflects the reality that many functions can share the same derivative, highlighting the importance of the constant of integration in providing the complete picture.