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Chapter 5: Problem 30

\(\int \frac{\cos x d x}{\sqrt{1+\sin x}}=\) (A) \(-\frac{1}{2}(1+\sin x)^{1 / 2}+C\) (B) \(\ln \sqrt{1+\sin x}+C\) (C) \(2 \sqrt{1+\sin x}+C\) (D) \(\ln |1+\sin x|+C\)

Short Answer

Expert verified
The answer is (C) \(2 \sqrt{1+\sin x}+C\).

Step by step solution

01

Identify the Substitution

We start by recognizing the form of the integrand. Notice that the derivative of \(1 + \sin x\) is \(\cos x\). This suggests using the substitution \(u = 1 + \sin x\). Then, \(du = \cos x \ dx\).
02

Substitute and Simplify

Substitute the identified \(u\) into the integral. The integral becomes \(\int \frac{1}{\sqrt{u}} \, du\).
03

Integrate with Respect to u

The integral \(\int \frac{1}{\sqrt{u}} \, du\) can be calculated as \(2\sqrt{u} + C\). This is a standard integral result for \(u^{-1/2}\).
04

Back-Substitute

Replace \(u\) back to the original variable \(x\). Since \(u = 1 + \sin x\), substitute back to get \(2\sqrt{1 + \sin x} + C\).
05

Match with Given Options

Compare the solved integral \(2\sqrt{1 + \sin x} + C\) with the options given. We find that it matches option (C).

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Key Concepts

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

Integration Techniques
Integration is a fundamental concept in calculus, serving as the reverse process of differentiation. It enables us to accumulate quantities, track areas under curves, and solve a wide variety of mathematical problems. There are several techniques available for solving integrals:

  • **Basic Integration**: Use basic integration formulas for simple expressions such as polynomials.
  • **Substitution Method**: Simplifies the integration process by transforming the expression into a more easily integrable form.
  • **Integration by Parts**: Useful when dealing with products of functions by reducing the integral to a simpler form.
  • **Partial Fraction Decomposition**: Breaks complex rational functions into simpler fractions to simplify integration.
These methods are essential tools in both pure and applied mathematics, providing different approaches and strategies to tackle integrals depending on the specific function being integrated.
U-Substitution
The substitution method, often called "u-substitution," is a powerful technique used in integral calculus to simplify integrations. This approach involves changing variables to make an integral more manageable to solve. The general idea starts with picking a substitution, which can turn a complicated integral into a simpler one.

Here are the steps for u-substitution:
  • Identify a part of the integral that could simplify the process when substituted with a new variable, denoted as "u".
  • Calculate the differential of "u", giving you "du" in terms of the original variable and differential.
  • Replace the original variable and differential in the integral with "u" and "du", transforming the integral.
  • Solve the transformed integral鈥攐ften using known rules and simpler expressions.
  • Back-substitute the original variables back to express the answer in terms of the initial variable.
In the original step-by-step solution, we used u-substitution by letting \( u = 1 + \sin x \). This carefully chosen substitution simplifies the integrand and transforms it into a form that is straightforward to integrate.
Trigonometric Integrals
Trigonometric integrals are integrals that involve trigonometric functions like \( \sin x \) and \( \cos x \). These functions often appear in calculus problems, especially in physics and engineering contexts due to their periodic nature.

There are several strategies to tackle trigonometric integrals:
  • **Standard Trigonometric Identities and Formulas**: Sometimes, using identities such as \( \sin^2 x + \cos^2 x = 1 \) or \( \sin(2x) = 2\sin x\cos x \) can simplify the integral.
  • **Substitution**: As demonstrated in the problem, substitution can simplify expressions that involve trigonometric functions by transforming them into more familiar algebraic forms.
  • **Power Reduction and Trigonometric Substitutions**: Techniques involving reducing powers of trigonometric functions or substituting with identities can help solve more complex integrals.
Trigonometric integrals require careful manipulation, but knowing various techniques and identities makes managing them much easier.
Definite Integral Evaluation
Definite integrals represent the accumulation of a quantity over a specific interval. Unlike indefinite integrals that yield a general function plus a constant, definite integrals provide a specific numerical value essential for applications like finding areas under curves.

Here is the typical process of evaluating definite integrals:
  • Set up the integral by identifying the limits of integration, which describe the interval over which you are integrating.
  • Use an appropriate integration technique to evaluate the integral in terms of an antiderivative.
  • Apply the **Fundamental Theorem of Calculus**, which connects the evaluation of definite integrals to antiderivatives. It involves calculating the difference between the values of the antiderivative at the upper and lower limits.
  • Calculate to get the numerical result, which is the exact area under the curve within the bounds specified.
In the solved exercise, although our focus was on indefinite integration using techniques like substitution and understanding the integral form, these concepts lay the groundwork for tackling definite integrals by providing the precise functions needed to evaluate over given intervals.

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Most popular questions from this chapter

\(\int \frac{\cos (\theta-1) d \theta}{\sin ^{2}(\theta-1)}=\) (A) \(2 \ln \sin |\theta-1|+C\) (B) \(-\csc (\theta-1)+C\) (C) \(-\cot (\theta-1)+C\) (D) \(\csc (\theta-1)+C\)

\(\int \frac{y-1}{y+1} d y=\) (A) \(y-2 \ln |y+1|+C\) (B) \(1-\frac{2}{y+1}+C\) (C) \(\ln \frac{|y|}{(y+1)^{2}}+C\) (D) \(1-2 \ln |y+1|+C\)

\(\int x^{3} \ln x d x=\) (A) \(x^{2}(3 \ln x+1)+C\) (B) \(\frac{x^{4}}{16}(4 \ln x-1)+C\) (C) \(\frac{x^{4}}{4}(\ln x-1)+C\) (D) \(3 x^{2}\left(\ln x-\frac{1}{2}\right)+C\)

\(\int(2-3 x)^{5} d x=\) (A) \(\frac{1}{6}(2-3 x)^{6}+C\) (B) \(-\frac{1}{2}(2-3 x)^{6}+C\) (C) \(\frac{1}{2}(2-3 x)^{6}+C\) (D) \(-\frac{1}{18}(2-3 x)^{6}+C\)

\(\int \frac{2 x+1}{2 x} d x=\) (A) \(x+\frac{1}{2} \ln |x|+C\) (B) \(x+2 \ln |x|+C\) (C) \(x+\ln |2 x|+C\) (D) \(\frac{1}{2}\left(2 x-\frac{1}{x^{2}}\right)+C\)
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