/*! 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 37 Evaluate the given integral. $... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 37

Evaluate the given integral. $$ \int \frac{x^{2}+1}{\left(9-x^{2}\right)^{3 / 2}} d x $$

Short Answer

Expert verified
\( \frac{10}{9} \cdot \frac{x}{\sqrt{9-x^2}} - \arcsin\left(\frac{x}{3}\right) + C \)

Step by step solution

01

Identify the Integral Type

Identify that the integral involves a rational function with a square root term in the denominator. This suggests that a trigonometric substitution might simplify the integral.
02

Choose an Appropriate Trigonometric Substitution

Use the substitution \( x = 3\sin\theta \), which means \( dx = 3\cos\theta \, d\theta \). This substitution simplifies \( 9 - x^2 = 9(1-\sin^2\theta) = 9\cos^2\theta \).
03

Substitute and Simplify the Integral

Substitute \( x = 3\sin\theta \) and \( dx = 3\cos\theta \, d\theta \) into the integral:\[ \int \frac{(3\sin\theta)^2 + 1}{(9 - (3\sin\theta)^2)^{3/2}} \cdot 3\cos\theta \, d\theta \]This becomes:\[ \int \frac{9\sin^2\theta + 1}{9^{3/2}\cos^3\theta} \cdot 3\cos\theta \, d\theta \]Simplify to:\[ \int \frac{(9\sin^2\theta + 1) \cdot 3}{27\cos^2\theta} \, d\theta \]
04

Further Simplify

The expression simplifies to:\[ \int \frac{9\sin^2\theta + 1}{9\cos^2\theta} \, d\theta = \int \left( \frac{\sin^2\theta}{\cos^2\theta} + \frac{1}{9\cos^2\theta} \right) \, d\theta \]Further simplify \( \frac{\sin^2\theta}{\cos^2\theta} \) to \( \tan^2\theta \):\[ \int \tan^2\theta \, d\theta + \frac{1}{9} \int \sec^2\theta \, d\theta \]
05

Integrate Each Component

Integrate \( \tan^2\theta \) by recognizing it as \( \sec^2\theta - 1 \):\[ \int \tan^2\theta \, d\theta = \int (\sec^2\theta - 1) \, d\theta = \int \sec^2\theta \, d\theta - \int 1 \, d\theta \]Which results in:\[ \tan\theta - \theta + C_1 \]Integrate \( \sec^2\theta \):\[ \frac{1}{9} \int \sec^2\theta \, d\theta = \frac{1}{9} \tan\theta + C_2 \]
06

Substitute Back and Complete the Integration

Recall the substitution \( x = 3\sin\theta \), so \( \sin\theta = \frac{x}{3} \) and \( \tan\theta = \frac{x}{\sqrt{9-x^2}} \). Substitute back:\[ \left(\frac{x}{\sqrt{9-x^2}}\right) - \arcsin\left(\frac{x}{3}\right) + \frac{1}{9} \left(\frac{x}{\sqrt{9-x^2}}\right) + C \]
07

Combine Results and Simplify

Combine like terms:\[ \frac{10}{9} \cdot \frac{x}{\sqrt{9-x^2}} - \arcsin\left(\frac{x}{3}\right) + C \]

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Key Concepts

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

Integral Calculus
Integral calculus is an essential part of calculus focused on the concept of integration, which is the process of finding the integral of a function. It essentially helps in determining the area under a curve, among other things. The integral \( \int \, f(x) \, dx \) can be thought of as the reverse operation of differentiation.When performing integration, one of the starting points is to identify the type of function you are dealing with. Is it polynomial, rational, trigonometric, or transcendental? Each function type might require a different technique for integration. For example, in this exercise, we are dealing with a rational function that includes a square root term in the denominator. This indicates that a trigonometric substitution could simplify the integral.Integral calculus is not only used for area calculations but also in physics for computing quantities like volume, mass, and electric charge, wherever summing up small quantities is needed."},{

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Most popular questions from this chapter

In each of Exercises \(31-40\), determine whether the given improper integral is convergent or divergent. If it converges, then evaluate it. \(\int_{0}^{1} \frac{1}{x \ln (x)} d x\)

In each of Exercises \(41-54,\) determine whether the given improper integral is convergent or divergent. If it converges, then evaluate it. \(\int_{-5}^{2} \ln (x+5) d x\)

A continuous function \(f(x)\) is given. Determine a function \(g(x)=c x^{p}\) such that (a) \(0 \leq f(x)\) \(\leq g(x)\) for each \(x\) in \([1, \infty),\) and (b) \(\int_{1}^{\infty} g(x) d x\) is convergent. This shows that \(\int_{1}^{\infty} f(x) d x\) is convergent by the Comparison Theorem. By determining a positive \(\varepsilon\) such that \(\int_{\varepsilon}^{\infty} g(x) d x<5 \times 10^{-3},\) approximate \(\int_{1}^{\infty} f(x) d x\) to two decimal places. $$ f(x)=x^{-6} \sqrt{1+3 x^{4}} $$

The curves \(y=\sin ^{2}(x) \cos ^{3}(x)\) and \(y=\sin ^{5}(x) \cos ^{2}(x)\) have a point of intersection with abscissa \(b\) in [0.5,1] . Find the area of the region between the two curves for \(0 \leq x \leq b\).

Show that the improper integral \(\Gamma(s)=\int_{0}^{\infty} x^{s-1} e^{-x} d x\) is convergent for \(s>0\). This function of \(s\) is called the gamma function.
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