/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Evaluate the integral. $$ \i... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the integral. $$ \int_{0}^{1 / \sqrt{2}} \frac{\arcsin x}{\sqrt{1-x^{2}}} d x $$

Short Answer

Expert verified
The result of the definite integral is 0.

Step by step solution

01

Identify the Integral Form

Recognize the integral form: \[\int\frac{f'(x)}{\sqrt{1-x^{2}}} dx = f(x)\sqrt{1-x^{2}} + C. \]where in our case \(f(x) = \arcsin x\). Hence, the integral \(f'(x)\) becomes \( \frac{1}{\sqrt{1-x^{2}}}\). So, our integral becomes a direct representation of this standard form.
02

Substitution and Integration

Replace \(\arcsin x\) according to the formula, so the integral becomes: \[\arcsin x \cdot \sqrt{1-x^{2}} + C. \]
03

Evaluate the definite integral

Next, substitute the limits from 0 to \(1 / \sqrt{2}\) into the integral to calculate the definite integral. It results in: \[ \left.\arcsin x \cdot \sqrt{1 - x^{2}}\right|_0^{1/\sqrt{2}}\]
04

Simplify the expression

Calculate the expression at 1/\(\sqrt2\) and then at 0. This simplifies to: \[ \left(\arcsin (1 / \sqrt{2}) \cdot \sqrt{1 - (1 / \sqrt{2})^{2}}\right) - \left(\arcsin 0 \cdot \sqrt{1 - 0^{2}}\right)\] which further simplifies to: \[ \left(\frac{\pi}{4} \cdot 0\right) - (0 \cdot 1) = 0 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
A definite integral is a fundamental concept in calculus, representing the area under a curve from one point to another along a function. Unlike an indefinite integral, which gives a general antiderivative, a definite integral computes a specific numerical value.
The process of evaluating a definite integral involves several steps:
  • Identifying the function to be integrated, known as the integrand.
  • Determining the limits of integration, which are the start and end points on the x-axis.
  • Computing the antiderivative, or the integral of the function.
  • Substituting the limits into the antiderivative and performing the subtraction to find the numerical result.
In this exercise, we dealt with the integral of \( \int_{0}^{1 / \sqrt{2}} \frac{\arcsin x}{\sqrt{1-x^{2}}} dx \). By evaluating the definite integral, we not only solve for the area but also confirm that the calculations bring us back to zero, owing to the canceling terms. Understanding the flow and purpose of each step is crucial in mastering calculus.
Substitution Method
The substitution method is an essential tool in calculus for simplifying complex integrals. It involves changing variables to transform a difficult integral into a more manageable one. This method revolves around the idea of 'undoing' a composite function, often simplifying integration by substitution with a simpler expression.
Here's how the substitution method generally works:
  • Identify a substitution that simplifies the integrand. A common approach is to let \( u = g(x) \) where \( g'(x) \) is part of the integrand.
  • Differentiate \( u \) to find \( du = g'(x) dx \).
  • Rewrite the integral in terms of \( u \) and \( du \), carrying through necessary algebraic adjustments.
  • Perform the integration on the simpler expression.
  • Convert back to the original variable after integration, if necessary.
In our exercise, recognizing \( f(x) = \arcsin x \) helped in substitution, as the derivative \( f'(x) = \frac{1}{\sqrt{1-x^2}} \) is a direct part of the integrand. This neat alignment allowed for seamless substitution, simplifying the integration process.
Trigonometric Integration
Trigonometric integration helps in tackling integrals involving trigonometric functions. This technique often requires recognizing trigonometric identities and relationships to transform integrals into solvable forms. In calculus, some integrals naturally involve trigonometric functions due to their periodic and oscillatory nature.
Some strategies for trigonometric integration include:
  • Using trigonometric identities like \( \sin^2 x + \cos^2 x = 1 \) to simplify expressions.
  • Applying substitutions linked with inverse trigonometric functions, such as \( x = \sin \theta \), which relate directly to the problem at hand.
  • Breaking down complex trigonometric expressions into simpler components that are easier to integrate.
In evaluating the integral \( \int \frac{\arcsin x}{\sqrt{1-x^{2}}} dx \), the function inside the square root, \( \sqrt{1-x^2} \), is trigonometric in nature, reminiscent of the identity for sine and cosine. Utilizing these trigonometric concepts allowed for the incorporation of arcsine and directly guided our integration process, ultimately leading us to find the solution.

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Most popular questions from this chapter

Show that \(\arctan (\sinh x)=\arcsin (\tanh x)\)

An object is projected upward from ground level with an initial velocity of 500 feet per second. In this exercise, the goal is to analyze the motion of the object during its upward flight. (a) If air resistance is neglected, find the velocity of the object as a function of time. Use a graphing utility to graph this function. (b) Use the result in part (a) to find the position function and determine the maximum height attained by the object. (c) If the air resistance is proportional to the square of the velocity, you obtain the equation $$ \frac{d v}{d t}=-\left(32+k v^{2}\right) $$ where \(-32\) feet per second per second is the acceleration due to gravity and \(k\) is a constant. Find the velocity as a function of time by solving the equation $$ \int \frac{d v}{32+k v^{2}}=-\int d t $$ (d) Use a graphing utility to graph the velocity function \(v(t)\) in part (c) if \(k=0.001\). Use the graph to approximate the time \(t_{0}\) at which the object reaches its maximum height. (e) Use the integration capabilities of a graphing utility to approximate the integral \(\int_{0}^{t_{0}} v(t) d t\) where \(v(t)\) and \(t_{0}\) are those found in part (d). This is the approximation of the maximum height of the object. (f) Explain the difference between the results in parts (b) and (e).

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