Evaluate The Integral. 3 5 Arctan 1 X Dx 1

Evaluating the Integral: ∫(3/5)arctan(1/x) dx from 1 to 3

Evaluating definite integrals often requires a blend of clever techniques and a solid understanding of calculus fundamentals. This article delves into the process of evaluating the definite integral: ∫(3/5)arctan(1/x) dx from 1 to 3. We'll explore various methods, discuss the challenges involved, and provide a comprehensive solution.

Understanding the Integral

The integral we're tasked with solving is:

(3/5) ∫₁³ arctan(1/x) dx

This integral presents a unique challenge because the antiderivative of arctan(1/x) isn't immediately obvious. We can't directly apply standard integration rules. This requires a strategic approach using integration techniques.

Strategy: Integration by Parts

A powerful technique for tackling integrals involving inverse trigonometric functions is integration by parts. This method is based on the product rule for differentiation and allows us to rewrite the integral into a more manageable form. The formula for integration by parts is:

∫u dv = uv - ∫v du

Applying Integration by Parts

To apply integration by parts to our integral, we need to carefully choose our 'u' and 'dv'. Let's define:

• u = arctan(1/x)
• dv = dx

Now, we need to find 'du' and 'v':

• du = d(arctan(1/x)) = (-1/(x²(1 + (1/x)²)) dx = (-1/(x² + 1)) dx (This requires the chain rule and the derivative of arctan(x)).
• v = ∫dx = x

Substituting these values into the integration by parts formula, we get:

(3/5) ∫arctan(1/x) dx = (3/5)[x * arctan(1/x) - ∫x * (-1/(x² + 1)) dx]

Solving the Remaining Integral

We now have a simpler integral to solve:

∫x * (-1/(x² + 1)) dx

This integral can be solved using substitution. Let's let:

• w = x² + 1
• dw = 2x dx

Therefore, x dx = (1/2) dw

Substituting, we get:

∫(-1/(x² + 1)) * x dx = ∫(-1/w) * (1/2) dw = (-1/2) ∫(1/w) dw = (-1/2) ln|w| + C

Back-Substitution and the Definite Integral

Substituting back for 'w', we have:

(-1/2) ln|x² + 1| + C

Now, we can put this back into our original integration by parts result:

(3/5) ∫arctan(1/x) dx = (3/5)[x * arctan(1/x) + (1/2) ln|x² + 1|] + C

Evaluating the Definite Integral

Now that we have the indefinite integral, we can evaluate the definite integral from 1 to 3:

(3/5)[x * arctan(1/x) + (1/2) ln|x² + 1|] |₁³

This means we substitute 3 and 1 into the expression and subtract the results:

(3/5) * [(3 * arctan(1/3) + (1/2) ln|3² + 1|) - (1 * arctan(1) + (1/2) ln|1² + 1|)]

(3/5) * [(3 * arctan(1/3) + (1/2) ln(10)) - (1 * π/4 + (1/2) ln(2))]

This simplifies to:

(3/5) * [3arctan(1/3) + (1/2)ln(10) - π/4 - (1/2)ln(2)]

(3/5) * [3arctan(1/3) + (1/2)ln(5) - π/4]

Numerical Approximation

Since arctan(1/3) is not a simple rational multiple of π, we need to use a calculator or numerical methods to approximate the value. Using a calculator:

arctan(1/3) ≈ 0.32175

Substituting this value:

(3/5) * [3(0.32175) + (1/2)ln(5) - π/4] ≈ (3/5) * [0.96525 + 0.8047 - 0.7854] ≈ (3/5) * [0.98455] ≈ 0.5907

Conclusion: A Step-by-Step Solution

The process of evaluating this integral involved multiple steps:

1. Identifying the appropriate technique: Integration by parts was the key to simplifying the integral.
2. Careful selection of 'u' and 'dv': The choice of 'u' and 'dv' significantly impacts the complexity of the resulting integral.
3. Solving the resulting integrals: We used substitution to solve the integral generated after applying integration by parts.
4. Back-substitution: We carefully substituted back to express the result in terms of the original variable 'x'.
5. Evaluating the definite integral: Substituting the limits of integration (1 and 3) and subtracting the results yielded the final numerical answer.
6. Numerical approximation: Due to the presence of arctan(1/3), a numerical approximation was necessary to obtain a final value.

Therefore, the value of the definite integral (3/5)∫₁³ arctan(1/x) dx is approximately 0.5907. This problem highlights the importance of choosing the right integration technique and performing each step meticulously to arrive at the correct solution. Remember that numerical approximations are sometimes necessary when dealing with transcendental functions. This detailed walkthrough provides a comprehensive understanding of how to handle similar integrals involving inverse trigonometric functions. The key is strategic application of integration by parts and careful attention to detail throughout the calculation.