Question

Find \(\frac{d}{d t} \int_{0}^{\sin t} \frac{\sin ^{-1} x}{x} d x.\)

Short answer

The derivative is \( t \cot t \).

Step-by-step solution

- Identify the composition

Recognize that the problem involves differentiating an integral where the limit of integration is a function of the variable of differentiation, specifically \( \sin t \). The integral can be expressed as \[ F(t) = \int_{0}^{\sin t} \frac{\sin^{-1} x}{x} dx \].

- Apply the Leibniz Rule

Apply the Leibniz Integral Rule, which states that for \( F(t) = \int_{a(t)}^{b(t)} f(x,t) dx \), its derivative is given as: \[ \frac{d}{dt} F(t) = f(b(t), t) \cdot b'(t) - f(a(t), t) \cdot a'(t), \] noting that \( a(t) = 0 \), \( b(t) = \sin t \), and \( f(x) = \frac{\sin^{-1} x}{x} \) with respect to \( x \), along with their respective derivatives with respect to \( t \).

- Evaluate the terms

Since \( a(t) = 0 \) is a constant, \( a'(t) = 0 \). Thus, the term \( f(a(t), t) \cdot a'(t) = 0 \). Now, evaluate \( f(b(t), t) = f(\sin t, t) = \frac{\sin^{-1}(\sin t)}{\sin t} \) and \( b'(t) = \cos t \).

- Simplify the expression

Since \( \sin^{-1}(\sin t) = t \) for \( t \) in the principal range \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), the function simplifies to \( \frac{t}{\sin t} \). Therefore, \[ \frac{d}{dt} \int_{0}^{\sin t} \frac{\sin^{-1} x}{x} dx = \frac{t}{\sin t} \cdot \cos t. \]

- Finalize the solution

Combine the simplified terms to arrive at the final derivative as \[ \frac{t \cos t}{\sin t} = t \cot t. \]

Key concepts

Leibniz Rule

The Leibniz Rule is a powerful tool in calculus, especially when dealing with the differentiation of integrals whose limits are functions of the variable of differentiation. In simple terms, it helps us find the derivative of an integral where the upper or lower limit is not constant. The rule can be expressed as follows:
For a function defined as \(F(t) = \int_{a(t)}^{b(t)} f(x,t) \, dx\), the derivative is given by: \( \frac{d}{dt} F(t) = f(b(t), t) \cdot b'(t) - f(a(t), t) \cdot a'(t)\).
This equation lets us break down and calculate each part easily. To use this, we need the values \(a(t)\) and \(b(t)\), their derivatives \(a'(t)\) and \(b'(t)\), and the function inside the integral \(f(x,t)\) evaluated at the limits. By applying this rule, we can handle complex integrals more easily.
Remember:

• Identify the variable limits of the integral.
• Evaluate the integrand at these limits.
• Find the derivatives of the limits with respect to the variable.

With practice, you'll see how powerful and straightforward Leibniz Rule makes differentiating integrals with variable limits!

Integration by parts

Integration by parts is another fundamental technique in calculus used for integrating products of functions. It's essentially the reverse of the product rule for differentiation. The formula for Integration by Parts is:
\( \int u \, dv = uv - \int v \, du \)
Where:

• \(u\) and \(dv\) are parts of the integrand.
• \(du\) is the derivative of \(u\).
• \(v\) is the integral of \(dv\).

Integration by parts works best when you can identify a part of the integrand that simplifies when differentiated (\(u\)), and another part that is easy to integrate (\(dv\)). Typically, we use the ILATE rule to choose \(u\) and \(dv\):

1. Inverse trigonometric functions (\(\sin^{-1} x, \cos^{-1} x\))
2. Logarithmic functions (\(ln x\))
3. Algebraic functions (\(x^2, x^3\))
4. Trigonometric functions (\(\sin x, \cos x\))
5. Exponential functions (\(e^x\))

Practice makes perfect. The more you work with this technique, the better you'll become at choosing the right parts and simplifying the integrals.

Inverse trigonometric functions

Inverse trigonometric functions are the inverse operations of the basic trigonometric functions. They allow us to find the angle that corresponds to a given trigonometric value. The primary inverse trigonometric functions are:

• \(\sin^{-1} x\) - arcsine
• \(\cos^{-1} x\) - arccosine
• \(\tan^{-1} x\) - arctangent
• \(\cot^{-1} x\) - arccotangent
• \(\sec^{-1} x\) - arcsecant
• \(\csc^{-1} x\) - arccosecant

Each of these functions returns an angle within a specific range. For example:

• \(\sin^{-1} x\) returns values in \([-\frac{\pi}{2}, \frac{\pi}{2}]\)
• \(\cos^{-1} x\) returns values in \([0, \pi]\)
• \(\tan^{-1} x\) returns values in \([-\frac{\pi}{2}, \frac{\pi}{2}]\)

Inverse trigonometric functions are used in many areas, including geometry, physics, and engineering. Understanding their properties is crucial for more advanced applications in calculus. They also frequently appear in integrals, making it essential to grasp how they are manipulated and differentiated. For example, in our exercise:
\(\sin^{-1} (\sin t) = t\) for \(t\) in \([-\frac{\pi}{2}, \frac{\pi}{2}]\), making computations simpler and straightforward.