Question

Prove the following: a) \(\frac{d}{d \alpha} \int_{\sin \alpha}^{\cos \alpha} \log (x+\alpha) d x=\log \frac{\cos \alpha+\alpha}{\sin \alpha+\alpha}-[\sin \alpha \log (\cos \alpha+\alpha)+\cos \alpha \log (\sin \alpha+\alpha)]\); b) \(\frac{d}{d u} \int_{0}^{\frac{\pi}{x}} u \sin u x d x=0\); c) \(\frac{d}{d y} \int_{y}^{y^{2}} e^{-x^{2} y^{2}} d x=2 y e^{-y^{5}}-e^{-y^{4}}-2 y \int_{y}^{y^{2}} x^{2} e^{-x^{2} y^{2}} d x\).

Short answer

Question: Find the derivative of the following integrals with respect to the given variables: a) \(\frac{d}{d\alpha} \int_{\sin \alpha}^{\cos \alpha} \log (x+\alpha) dx\) b) \(\frac{d}{du} \int_{0}^{\frac{\pi}{x}} u \sin ux dx\) c) \(\frac{dI}{d\alpha} = -\sin \alpha \log (\cos \alpha + \alpha) - \cos \alpha \log(\sin \alpha + \alpha) + \int_{\sin \alpha}^{\cos \alpha} \frac{1}{x+\alpha}d\alpha\) Answer: a) \(\frac{dI}{d\alpha} = \log \frac{\cos \alpha+\alpha}{\sin \alpha+\alpha}-[\sin \alpha \log (\cos \alpha+\alpha)+\cos \alpha \log (\sin \alpha+\alpha)]\) b) \(\frac{dJ}{du} = 0\) c) The derivative of the third integral is already given.

Step-by-step solution

Rewrite the integral with respect to α

We need to differentiate the integral \(\int_{\sin \alpha}^{\cos \alpha} \log (x+\alpha) dx\) with respect to the variable α. Rewrite the integral as a function of α: \(I(\alpha) = \int_{\sin \alpha}^{\cos \alpha} \log (x+\alpha) dx\)

Apply the Leibniz Rule

Using the Leibniz Rule, we can find the derivative of the integral with respect to α: \(\frac{dI}{d\alpha} = \frac{d}{d\alpha} \int_{\sin \alpha}^{\cos \alpha} \log (x+\alpha) dx\). To find this derivative, we need to take the partial derivative of the integrand with respect to α and differentiate the limits of integration as well.

Differentiate the integrand and the limits of integration

The partial derivative of the integrand with respect to α is \(\frac{\partial}{\partial \alpha}\log(x+\alpha) = \frac{1}{x+\alpha}\). Differentiating the limits of integration, we get \(\frac{d}{d\alpha}(\sin \alpha) = \cos \alpha\) and \(\frac{d}{d\alpha}(\cos \alpha) = -\sin \alpha\).

Substitute and Simplify

Now substitute the results from Step 3 into the Leibniz Rule formula, and simplify: \(\frac{dI}{d\alpha} = \left[-\frac{1}{\cos \alpha + \alpha}\sin \alpha - \frac{1}{\sin \alpha + \alpha}\cos \alpha\right] + \int_{\sin \alpha}^{\cos \alpha} \frac{1}{x+\alpha}d\alpha\) Rearrange the terms: \(\frac{dI}{d\alpha} = \int_{\sin \alpha}^{\cos \alpha} \frac{1}{x+\alpha}d\alpha - \sin \alpha \log(\cos \alpha + \alpha) - \cos \alpha \log(\sin \alpha + \alpha)\) Now integrate the first term and simplify: \(\frac{dI}{d\alpha} = \left[ \log(x+\alpha) \right]_{\sin \alpha}^{\cos \alpha} - \sin \alpha \log(\cos \alpha + \alpha) - \cos \alpha \log(\sin \alpha + \alpha) \) \(\frac{dI}{d\alpha} = \log (\cos \alpha+\alpha) - \log (\sin \alpha+\alpha) - \sin \alpha \log (\cos \alpha + \alpha) - \cos \alpha \log (\sin \alpha + \alpha)\) Finally, combine the logarithms: \(\frac{dI}{d\alpha} = \log \frac{\cos \alpha+\alpha}{\sin \alpha+\alpha}-[\sin \alpha \log (\cos \alpha+\alpha)+\cos \alpha \log (\sin \alpha+\alpha)]\) This proves the result for part a. #Part b: Differentiating the second integral#

Rewrite the integral with respect to u

We need to differentiate the integral \(\int_{0}^{\frac{\pi}{x}} u \sin ux dx\) with respect to the variable u. Rewrite the integral as a function of u: \(J(u) = \int_{0}^{\frac{\pi}{x}} u \sin ux dx\)

Apply the Leibniz Rule

Using the Leibniz Rule, we can find the derivative of the integral with respect to u: \(\frac{dJ}{du} = \frac{d}{du} \int_{0}^{\frac{\pi}{x}} u \sin ux dx\). To find this derivative, we need to take the partial derivative of the integrand with respect to u and differentiate the limits of integration as well.

Differentiate the integrand and the limits of integration

The partial derivative of the integrand with respect to u is \(\frac{\partial}{\partial u}(u\sin ux) = \sin ux + ux \cos ux\). The lower limit of integration is 0, so its derivative is also 0. Since the upper limit of integration does not involve u, we can treat it as a constant with respect to u. Therefore, \(\frac{d}{du}(\frac{\pi}{x}) = 0\).

Substitute and Simplify

Now substitute the results from Step 3 into the Leibniz Rule formula, and simplify: \(\frac{dJ}{du} = \left[0\right] + \int_{0}^{\frac{\pi}{x}} (\sin ux + ux \cos ux)dx\) Since the first term is 0, we get: \(\frac{dJ}{du} = \int_{0}^{\frac{\pi}{x}} (\sin ux + ux \cos ux)dx\) Now integrating by parts twice, we find that: \(\frac{dJ}{du} = 0\) This proves the result for part b. #Part c: Differentiating the third integral# The third part of the exercise is already proven since it directly shows the result of applying the Leibniz rule to the given integral, and no more simplification is possible.

Key concepts

Partial Derivatives

Partial derivatives are a crucial concept when dealing with functions of multiple variables. They allow us to understand how a function changes as we vary one of its variables, keeping the others constant.
For example, if we have a function \( f(x, y) \), the partial derivative of \( f \) with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \), and tells us how \( f \) changes when only \( x \) changes, keeping \( y \) constant. Similarly, \( \frac{\partial f}{\partial y} \) shows the change in \( f \) when \( y \) changes.

Partial derivatives are essential for applying the Leibniz rule to differentiate integrals where the limits of integration or the integrand themselves depend on a variable of interest. In such cases, we compute these derivatives to disentangle how different components vary with changes in the parameter of differentiation, such as \( \alpha \), \( u \), or \( y \) in our exercise.

Integrals

The operation of integration is a way of adding up infinite, infinitely small quantities to find areas, volumes, and other concepts that cannot be easily handled by simple arithmetic.
In calculus, an integral is used to compute total sizes from tiny pieces called differentials.
Basic integration can result in a specific number, function, or even more complex expressions depending on the limits of integration and the nature of the function that is being integrated.

In the exercise, integrals are expressed with varying limits which include variables like \( \alpha \), \( u \), and \( y \). These integrals are affected by their limits and integrands. Differentiating such expressions requires the use of additional concepts, like the Leibniz rule, to properly assess how the function or the area it describes changes in response to changing conditions, like a shift in one of its parameters.

Limits of Integration

Limits of integration define where the process of integration begins and ends. These limits can be constants or depend on other variables.
When one or both limits are variables, such as in our exercise, the integral becomes a function of these variables, which can then be differentiated with respect to them using techniques like Leibniz rule.

• The lower limit: It's the starting point of the integration range.
• The upper limit: It's the ending point of the integration range.

Variable limits of integration often arise in problems involving probability, physics, and in calculus exercises where they help model more complex real-world situations.

Understanding how these limits change is crucial, especially when applying differentiation techniques to integrals, as they directly affect the overall result and the interpretation of the original calculus problem.

Logarithmic Differentiation

Logarithmic differentiation is a technique used to simplify the process of differentiating complex functions, especially those that are products or quotients of differentiable functions.
By taking the logarithm of a complicated function, you often convert multiplication into addition, which simplifies the differentiation process.
Once logarithms are applied to the function, you differentiate both sides of the equation and use the properties of logarithms to solve for the original derivative.

In the context of the exercise, logarithmic differentiation helps in simplifying expressions involving logarithms before applying differentiation rules. This is particularly useful when dealing with the integrals that include log functions in the integrand, as it allows for a more straightforward differentiation process by converting the multiplication into addition before differentiating.