Problem 365 - Calculus Volume 1

Question

In the following exercises, integrate using the indicated substitution. $$\int \ln (x) \frac{\sqrt{1-(\ln x)^{2}}}{x} d x ; u=\ln x$$

Step-by-Step Solution

Answer

\( \int \ln(x) \frac{\sqrt{1-(\ln x)^2}}{x} dx \) is solved using substitution \( u=\ln x \), partial integration, and trigonometric substitution.

Step 1: Identify the Substitution

We are given the substitution \( u = \ln x \). This indicates that we should express the integral in terms of \( u \) instead of \( x \).

Step 2: Differentiate the Substitution

Compute the derivative of \( u \) with respect to \( x \). We have \( u = \ln x \) which implies \( \frac{du}{dx} = \frac{1}{x} \). Therefore, \( du = \frac{1}{x} dx \).

Step 3: Change Variables in the Integral

Substitute \( u = \ln x \) and \( dx = x \, du \) into the integral. This gives \( \int \ln(x) \frac{\sqrt{1-(\ln x)^2}}{x} dx = \int u \sqrt{1-u^2} \, du \).

Step 4: Identify the New Integral

The integral has now been transformed into \( \int u \sqrt{1-u^2} \, du \). This is a new integral in terms of \( u \).

Step 5: Use Integration by Parts

We use integration by parts, with \( v = u \) and \( dw = \sqrt{1-u^2} \, du \). Differentiate and integrate the respective parts: \( dv = du \) and \( w = \int \sqrt{1-u^2} \, du \) which requires a trigonometric substitution.

Step 6: Solve \( \int \sqrt{1-u^2} \, du \) Using Trig Substitution

Set \( u = \sin(t) \), so \( du = \cos(t) dt \). This transforms the integral to \( \int \sqrt{1-u^2} \, du = \int \sqrt{1-\sin^2(t)} \cos(t) \, dt = \int \cos^2(t) \, dt \).

Step 7: Integrate \( \cos^2(t) \) Using Double Angle Identity

Use the identity \( \cos^2(t) = \frac{1+\cos(2t)}{2} \) to simplify the integral: \( \int \cos^2(t) \, dt = \int \frac{1+\cos(2t)}{2} \, dt \). This becomes \( \frac{t}{2} + \frac{1}{4}\sin(2t) + C \).

Step 8: Substitute Back to \( u \)-terms

Convert back to \( u \) using \( u = \sin(t) \), so \( \arcsin(u) = t \). Therefore, \( \int \sqrt{1-u^2} \, du = \frac{1}{2} \arcsin(u) + \frac{1}{4} u\sqrt{1-u^2} + C \).

Step 9: Complete Integration by Parts Formula

Using integration by parts \( \int u \sqrt{1-u^2} \, du = u \times \left( \frac{1}{2} \arcsin(u) + \frac{1}{4} u\sqrt{1-u^2} \right) - \int \sqrt{1-u^2} \, du \). Simplify to get the final formula.

Step 10: Substitute Back to \( x \)-variables

Change back from \( u \) to \( x \) with \( u = \ln x \). Substitute this back into your final integration result to complete solving the integral.

Key Concepts

Substitution Method

The substitution method is a powerful technique in integral calculus used to simplify complex integrals by changing variables. This approach focuses on finding a substitution that converts the integral into a simpler form.
In the original exercise, the substitution given was \( u = \ln x \). This allowed us to transform the integral from one in terms of \( x \) to one in terms of \( u \).
Here's how the substitution mechanism works:

• Identify a part of the integrand to substitute, which simplifies the integral when replaced by a single variable \( u \).
• Express the differential \( dx \) in terms of \( du \), using \(\frac{du}{dx} = \frac{d}{dx}\ln x = \frac{1}{x} \), resulting in \( du = \frac{1}{x} dx \).
• Replace all instances of the original variable and \( dx \) in the integral with the new variable \( u \) and differential \( du \).

This method simplifies the original expression, paving the way for further integration techniques such as integration by parts or trigonometric substitution.

Integration by Parts

Integration by parts is another essential technique in integral calculus, particularly useful for integrals involving the product of two functions. It is essentially the reverse process of the product rule in differentiation and is given by the formula: \[ \int u \, dv = uv - \int v \, du \]To apply this method, you should:

• Select \( u \) and \( dv \) from the integral such that differentiating \( u \) and integrating \( dv \) simplifies the problem.
• Differentiate \( u \) to find \( du \) and integrate \( dv \) to find \( v \).
• Substitute these values into the integration by parts formula.

In our exercise after substitution, the integral became \( \int u \sqrt{1-u^2} \, du \). Applying integration by parts involved choosing \( v = u \) and \( dw = \sqrt{1-u^2} \, du \). The solution then required trigonometric substitution to further evaluate \( \int \sqrt{1-u^2} \, du \).

Trigonometric Substitution

Trigonometric substitution is used to simplify integrals containing square roots by leveraging trigonometric identities. This method often solves integrals of the form \( \sqrt{a^2 - u^2} \), \( \sqrt{u^2 - a^2} \), and \( \sqrt{u^2 + a^2} \).
For the integral \( \int \sqrt{1-u^2} \, du \) in our problem, trigonometric substitution was key:

• Set \( u = \sin(t) \), which implies \( du = \cos(t) dt \).
• This transforms the integral into \( \int \cos^2(t) \, dt \).
• Employ trigonometric identities like \( \cos^2(t) = \frac{1+\cos(2t)}{2} \) to simplify and solve the integral.

Finally, after solving the integral in terms of \( t \), revert back to the original variable \( u \) using inverse trigonometric functions resulting from the substitution.

Integral Calculus

Integral calculus is a fundamental area of mathematics concerned with finding the integral or antiderivative of functions. It complements differential calculus, allowing us to compute areas, volumes, and other concepts related to sums and distances.
The key essence of integral calculus involves:

• Definite Integrals: Calculating the area under a curve between two points, represented by bounds.
• Indefinite Integrals: Finding a function whose derivative is the given function, the general antiderivative including a constant of integration \( C \).

In this exercise, we have primarily dealt with an indefinite integral, applying sophisticated techniques like substitution, integration by parts, and trigonometric substitution. These tools are vital in handling complex integrals encountered in real-world applications and advanced mathematics. Understanding and mastering these methods are key to unlocking the full potential of integral calculus.