Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int \arctan 5 x d x $$

Step-by-Step Solution

Answer

The integral is \( x \arctan(5x) - \frac{1}{10} \ln(1+(5x)^2) + C \).

Step 1: Choose u and dv

For the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we need to choose values for \( u \) and \( dv \). We choose \( u = \arctan(5x) \) because its derivative is simpler, and \( dv = dx \), which is easy to integrate.

Step 2: Differentiate u

Differentiate \( u = \arctan(5x) \) to find \( du \). The derivative is \( du = \frac{5}{1+(5x)^2} dx \).

Step 3: Integrate dv

Integrate \( dv = dx \) to find \( v \). The integral is straightforward: \( v = x \).

Step 4: Substitute into Integration by Parts Formula

Substitute the values of \( u, du, v, \) and \( dv \) into the integration by parts formula \( \int u \, dv = uv - \int v \, du \). So we have: \( \int \arctan(5x) \, dx = x \arctan(5x) - \int x \frac{5}{1+(5x)^2} dx \).

Step 5: Simplify and Complete the Integral

Simplify the integrand \( \int x \frac{5}{1+(5x)^2} dx \). Substitute \( u = 1 + (5x)^2 \), then \( du = 50x \, dx \) or \( \. du/50 = x \, dx \). Therefore, \( \int x \frac{5}{1+(5x)^2} dx = \frac{5}{50} \int \frac{1}{u} du = \frac{1}{10} \ln|u| + C' = \frac{1}{10} \ln(1+(5x)^2) + C' \).

Step 6: Finalize the Solution

Substitute back to get the final result: \( x \arctan(5x) - \frac{1}{10} \ln(1+(5x)^2) + C \), where \( C \) is the constant of integration.

Key Concepts

Definite Integrals

When it comes to calculus, definite integrals are a fundamental concept. They provide a way to find the accumulation of values over a specific interval. An exciting and practical application is determining the area under a curve within given limits.
For instance, suppose you have a continuous function, say, Mary's savings account over time. The definite integral allows you to calculate her total savings between two dates. You define this function over an interval, and the integral effectively sums up the small areas (representing daily savings) across this range.
Mathematically, if you have a function \( f(x) \) and you want to find the total "amount" between two points \( a \) and \( b \), you use the definite integral:

• Notation: \( \int_{a}^{b} f(x) \, dx \)
• The "\( a \)" and "\( b \)" are the limits of integration.

Remember, the result of a definite integral is a number, not a function, and this is incredibly helpful in solving real-world problems.

Trigonometric Integration

Trigonometric integration is another fascinating branch of calculus. It deals with integrating functions that involve trigonometric functions, like sine, cosine, tangent, and their inverses. You'll often apply strategies like substitution and using trigonometric identities to simplify these integrals.
For example, say you need to integrate \( \,\int \sin{x} \, dx \). You know the integral of sine is a basic one, resulting in \(-\cos{x} + C\). Simple enough, but it becomes more complex with products or powers.
Here are some tips:

• Use trigonometric identities to simplify before integrating.
• For products like \( \sin{x} \cos{x} \), consider using substitution or identities to transform it into a more accessible form.
• Inverse trigonometric functions might require integration by parts or specific inverses' derivatives.

Integrating trigonometric functions often blends different techniques for a solution.

Arc Functions

Arc functions, or inverse trigonometric functions, are closely tied to angles and their representations. The arc functions, such as \( \arcsin{x} \), \( \arccos{x} \), and of particular interest here, \( \arctan{x} \), return the angle whose trigonometric ratio is the input value. These are essential in calculus, especially for integration tasks involving inverse trig functions.
When working with inverse trigonometric functions:

• Each function has a specific derivative which can be memorized for quicker computation.
• For \( \arctan{x} \), the derivative is \( \frac{1}{1+x^2} \). This is vital for integration, considering integration by parts.
• These functions often pair well with integration techniques to solve integrals like \( \int \arctan{5x} \, dx \).

Knowing how to handle these functions can be immensely rewarding, allowing for deeper exploration of calculus challenges.