Question

In the following exercises, compute each integral using appropriate substitutions. \(\int \frac{d t}{t \sqrt{1-\ln ^{2} t}}\)

Short answer

\( \arcsin(\ln t) + C \) where \( C \) is the integration constant.

Step-by-step solution

Identify a Suitable Substitution

We notice that the expression inside the square root is of the form \(1 - \ln^2 t\). This suggests a substitution involving \(\ln t\). Let \(u = \ln t\). This implies that \(du = \frac{1}{t} dt\), or equivalently \(dt = t \, du\). Since \(t = e^u\), the differential becomes \(dt = e^u \, du\).

Rewrite the Integral in Terms of the New Variable

Substitute \(u = \ln t\) into the integral. The integral becomes:\[\int \frac{1}{t \sqrt{1 - \ln^2 t}} \, dt = \int \frac{1}{e^u \sqrt{1 - u^2}} \, e^u \, du = \int \frac{1}{\sqrt{1 - u^2}} \, du.\]

Identify the New Integral Form

The integral \(\int \frac{1}{\sqrt{1 - u^2}} \, du\) corresponds to the standard form of the inverse sine function derivative. Recall the integral formula:\[\int \frac{1}{\sqrt{1 - x^2}} \, dx = \arcsin(x) + C.\]

Compute the Integral Using the Standard Form

Using the standard inverse sine integral formula, the integral simplifies to:\[\int \frac{1}{\sqrt{1 - u^2}} \, du = \arcsin(u) + C,\]where \(C\) is the constant of integration.

Substitute Back the Original Variable

Recall that \(u = \ln t\). Thus, substituting back gives:\[\arcsin(u) + C = \arcsin(\ln t) + C.\]This is the final form of the integral.

Key concepts

Substitution Method

In calculus, when faced with complex integrals, the substitution method is a powerful technique to simplify the process. It involves changing the variable to transform the integral into an easier form. In the given exercise, we encountered the expression \(1 - \ln^2 t\), leading us to set \(u = \ln t\). This substitution simplifies the integral, as the differential \(du\) corresponds to \(\frac{1}{t} dt\), making computations more manageable.

The steps are straightforward:

• Select a new variable \(u\) (in this case, \(\ln t\)) to structure the transformation.
• Rewrite the differential \(dt\) in terms of \(u\) and \(du\).
• Reformulate the integral in terms of \(u\), allowing the complicated parts to become simpler.

Remember, mastering substitution gives you a tool to tackle a wide array of integrals, where recognizing forms is key to success.

Inverse Trigonometric Functions

Integral computations sometimes lead to expressions resembling inverse trigonometric functions. In this exercise, after the substitution, the new integral \(\int \frac{1}{\sqrt{1 - u^2}} \, du\) emerged.

This closely resembles the derivative of the inverse sine function, known as \(\arcsin x\). The standard integral form: \[\int \frac{1}{\sqrt{1 - x^2}} \, dx = \arcsin(x) + C\]is instrumental in recognizing such forms within integrals.

• Recognize the pattern following the form.
• Use inverse trigonometric identities to simplify integration tasks.
• Cast back to the original variable after finding the integral in terms of \(u\).

Inverse trigonometric functions bridge gaps in integral calculus by allowing elegant solutions for otherwise convoluted integral forms.

Integral Computation

Integral computation is at the core of calculus, involving finding the antiderivative or the area under a curve. The process can vary depending on the function's complexity. This exercise demonstrated that substitution can convert an integral into simpler forms, ultimately by recognizing a standard integral.

Key steps in integral computation include:

• Identify substitutions that transform the integral into a recognizable standard form.
• Recognize and apply known integral formulas, such as those for inverse trigonometric functions.
• Simplify and solve, always incorporating the constant of integration \(C\).

Each integral solved builds a deeper understanding of the function behavior, preparing you to tackle more complex forms effectively.

Differential Equations

Differential equations often require solving integrals as part of their solutions. These equations involve relationships between functions and their derivatives, essential in modeling real-world phenomena.

The techniques used in integral computation, such as substitution and recognizing inverse trigonometric forms, also aid in solving differential equations:

• Transform equations into a simpler form through strategic substitutions.
• Detect patterns and apply solutions from standard integral forms.
• Translate back to the original variable after solving the integral portion.

Integrals and differential equations frequently intertwine in calculus. Developing a strong grasp of integral methods enhances your ability to solve differential equations efficiently, crucial for advanced fields like physics and engineering.