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Chapter 5: Problem 398

In the following exercises, find each indefinite integral, using appropriate substitutions. $$\int \frac{d x}{\sqrt{1-16 x^{2}}}$$

Short Answer

Expert verified
\( \int \frac{d x}{\sqrt{1-16 x^{2}}} = \frac{1}{4} \arcsin(4x) + C \)

Step by step solution

01

Recognize Structure for Trigonometric Substitution

The integrand has the form \( \frac{1}{\sqrt{1 - 16x^2}} \), which suggests using the trigonometric substitution \( x = \frac{1}{4} \sin(\theta) \). This is because \( 1 - 16x^2 \) relates to the identity \( 1 - \sin^2(\theta) = \cos^2(\theta) \).
02

Apply Substitution

Substitute \( x = \frac{1}{4} \sin(\theta) \) into the integrand, thus \( dx = \frac{1}{4} \cos(\theta) d\theta \). Replace in the integral which becomes:\[\int \frac{\frac{1}{4} \cos(\theta) d\theta}{\sqrt{1 - 16(\frac{1}{4} \sin(\theta))^2 }} = \int \frac{\frac{1}{4} \cos(\theta) d\theta}{\sqrt{1 - \sin^2(\theta)}}\]
03

Simplify the Integral

Recognize that \( \sqrt{1 - \sin^2(\theta)} = \cos(\theta) \). Hence, the integrand simplifies:\[\int \frac{\frac{1}{4} \cos(\theta) d\theta}{\cos(\theta)} = \int \frac{1}{4} d\theta\]
04

Integrate with Respect to Theta

Integrate \( \int \frac{1}{4} d\theta = \frac{1}{4} \theta + C \), where \( C \) is the constant of integration.
05

Back-Substitute to Original Variable

Recall that \( x = \frac{1}{4} \sin(\theta) \) implies \( \theta = \arcsin(4x) \). Substitute back to get the integral in terms of \( x \):\[\frac{1}{4} \theta + C = \frac{1}{4} \arcsin(4x) + C\]

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Key Concepts

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

Trigonometric Substitution
Trigonometric substitution is a powerful technique used in integral calculus, especially for integrals involving square roots. The idea is to simplify the integrand by substituting a trigonometric function for the variable. In this example, the integral \( \int \frac{d x}{\sqrt{1-16 x^{2}}} \) suggests a trigonometric substitution because the expression under the square root resembles the Pythagorean identity \( 1 - \sin^2(\theta) = \cos^2(\theta) \).
  • We set \( x = \frac{1}{4} \sin(\theta) \), transforming \( dx = \frac{1}{4} \cos(\theta) d\theta \).
  • This substitution leverages the identity \( 1 - 16x^2 \rightarrow 1 - \sin^2(\theta) \), which simplifies to \( \cos^2(\theta) \).
By using trigonometric substitution, complex square roots are converted into simpler trigonometric identities, making integration more approachable.
Integral Calculus
Integral calculus is the branch of mathematics concerned with finding antiderivatives or integrals. Indefinite integrals, represented by \( \int f(x) \, dx \), find a function whose derivative is the integrand. In this exercise, our task was to integrate \( \int \frac{d x}{\sqrt{1-16 x^{2}}} \), a classic example for utilizing advanced techniques.
  • The primary goal is to remove the challenging square root in the denominator by a suitable substitution.
  • After choosing the right trigonometric substitution, the integral simplifies greatly, enabling ease of integration.
Integral calculus is not just about solving integrals; it is about choosing the right strategies that simplify these challenges, leading to an antiderivative effectively.
Substitution Method
The substitution method in calculus is essential for transforming an integral into a more workable form. This technique replaces an integral variable with a new one, reducing the complexity. In our scenario, the integral \( \int \frac{d x}{\sqrt{1-16 x^{2}}} \) was transformed by setting \( x = \frac{1}{4} \sin(\theta) \).
  • Upon substituting, \( dx \) translates to \( \frac{1}{4} \cos(\theta) d\theta \), and the denominator simplifies to \( \cos(\theta) \).
  • This simplification leads to a much easier integral \( \int \frac{1}{4} d\theta \).
The substitution method not only simplifies computation but also corresponds to transforming the integration problem into a basic, more solvable form. This is why it is deeply valued in the realm of calculus.

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Most popular questions from this chapter

Suppose that \(\int_{0}^{4} f(x) d x=5\) and \(\int_{0}^{2} f(x) d x=-3,\) and \(\int_{0}^{4} g(x) d x=-1\) and \(\int_{0}^{2} g(x) d x=2 .\) In the following exercises, compute the integrals. \(\int_{2}^{4}(f(x)-g(x)) d x\)

Find the area under the graph of \(f(t)=\frac{t}{\left(1+t^{2}\right)^{a}}\) between \(t=0\) and \(t=x\) where \(a>0\) and \(a \neq 1\) is fixed, and evaluate the limit as \(x \rightarrow \infty\) .

If \(f\) is 1 -periodic \((f(t+1)=f(t)), \quad\) odd, and integrable over \([0,1], \quad\) is it always true that \(\int_{0}^{1} f(t) d t=0 ?\)

In the following exercises, evaluate the indefinite integral \(\int f(x) d x\) with constant \(C=0\) using u-substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of \(C\) that would need to be added to the antiderivative to make it equal to the definite integral \(F(x)=\int_{a}^{x} f(t) d t,\) with a the left endpoint of the given interval. $$ \int \frac{\sin x}{\cos ^{3} x} d x \text { over }\left[-\frac{\pi}{3}, \frac{\pi}{3}\right] $$

In the following exercises, compute the average value using the left Riemann sums \(L_{N}\) for \(N=1,10,100\) . How does the accuracy compare with the given exact value? \([\mathrm{T}] \quad y=x e^{x^{2}}\) over the interval \([0,2] ;\) the exact solution is \(\frac{1}{4}\left(e^{4}-1\right)\)
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