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What change of variables is suggested by an integral containing \(\sqrt{100-x^{2}} ?\)

Short Answer

Expert verified
Answer: The trigonometric substitution \(x = 10\sin(\theta)\) can be used to simplify the expression \(\sqrt{100-x^2}\), turning it into a simpler trigonometric expression, \(10\cos(\theta)\).

Step by step solution

01

Identify the substitution

The expression inside the square root is \(100-x^2\). To simplify this, we can use a trigonometric substitution. A common substitution for this form is \(x=10\sin(\theta)\), where the constant 10 comes from the square root of 100.
02

Rewrite the expression with the substitution

Now, rewrite the given expression using the substitution \(x=10\sin(\theta)\): \(\sqrt{100-(10\sin(\theta))^2}\)
03

Simplify the expression

By simplifying the expression, we get: \(\sqrt{100-100\sin^2(\theta)} = \sqrt{100(1-\sin^2(\theta))}\) Using the Pythagorean identity \(1-\sin^2(\theta)=\cos^2(\theta)\), we can further simplify the expression: \(\sqrt{100\cos^2(\theta)}=10\cos(\theta)\) Now, the given expression has been converted to a simpler trigonometric expression by using the change of variables, \(x=10\sin(\theta)\).

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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 a fundamental component of mathematics concerned with finding the quantity where the rate of change is known. This process is complementary to differentiation, where we find the rate of change. In this context, the integral represents the area under a curve on a graph, often referred to as the antiderivative. When faced with an integral containing a complex expression like \(\sqrt{100-x^{2}}\), we're essentially looking to find the original function before it was differentiated and its graph became the curve we're now examining. The challenge is in simplifying the expression under the integral to make it more manageable for computation. Here, trigonometric substitution is a powerful technique helping us convert algebraic expressions into trigonometric ones, which can be more straightforward to integrate due to the library of known integrals for trigonometric functions. As you'll see in subsequent sections, specific substitutions streamline the integration process and make finding the solution a more intuitive endeavor.
Substitution Method
The substitution method in integral calculus is a powerful tool for simplifying difficult integrals. Think of it as a mathematical 'alchemical process' that transforms complex expressions into simpler ones, making them easier to manage and solve. In practice, it involves identifying a part of the integral that can be replaced with a variable to streamline the integration process. It's like swapping out a complex puzzle piece with one that fits more naturally into our existing puzzle. Once we replace the complicated expression with a simpler one, using a well-chosen substitution, we can often perform the integration much more easily. Then, we substitute back the original variables to find our solution. This reversal is essential to ensure that the final answer is in terms of the original variables.
Pythagorean Identity
The Pythagorean identity is an indispensable tool derived from the Pythagorean theorem, one of the cornerstones of trigonometry. This identity gives a relationship between the sine and cosine functions: \(1-\sin^2(\theta)=\cos^2(\theta)\). It's like having a secret passcode that unlocks a simpler form of an expression involving sine and cosine. In the process of integral calculus, it allows us to transform the squared sine function into a squared cosine function, or vice versa, which often simplifies the expression drastically. Recognizing when to employ this identity is key, as in the given example with \(\sqrt{100-x^{2}}\), where this identity helps collapse the expression into one involving a single trigonometric function, making it far more tractable.
Square Root Simplification
The square root simplification is key in streamlining complex mathematical expressions. It's like reducing a tough knot to a straight line, making it easier to work with. When dealing with integrals, simplifying square roots often leads to forms that are more amenable to standard integration techniques. For instance, \(\sqrt{100 - x^2}\) seems daunting at first, but with the right approach, it can be simplified using trigonometric substitution. By recognizing the structure of the term under the square root sign as a difference of squares, a common geometric scenario in trigonometry, we can simplify the expression. After applying the Pythagorean identity, the root sign can often be eradicated, leaving behind a more straightforward expression involving trigonometric functions, which are generally easier to integrate due to their periodic and well-understood properties.

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Most popular questions from this chapter

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. Suppose \(\int_{a}^{b} f(x) d x\) is approximated with Simpson's Rule using \(n=18\) subintervals, where \(\left|f^{(4)}(x)\right| \leq 1\) on \([a, b]\) The absolute error \(E_{S}\) in approximating the integral satisfies \(E_{s} \leq \frac{(\Delta x)^{5}}{10}\) 1\. If the number of subintervals used in the Midpoint Rule is increased by a factor of \(3,\) the error is expected to decrease by a factor of \(8 .\) c. If the number of subintervals used in the Trapezoid Rule is increased by a factor of \(4,\) the error is expected to decrease by a factor of \(16 .\)

The gamma function is defined by \(\Gamma(p)=\int_{0}^{\infty} x^{p-1} e^{-x} d x,\) for \(p\) not equal to zero or a negative integer. a. Use the reduction formula $$\int_{0}^{\infty} x^{p} e^{-x} d x=p \int_{0}^{\infty} x^{p-1} e^{-x} d x \quad \text { for } p=1,2,3, \ldots$$ to show that \(\Gamma(p+1)=p !(p\) factorial). b. Use the substitution \(x=u^{2}\) and the fact that \(\int_{0}^{\infty} e^{-u^{2}} d u=\frac{\sqrt{\pi}}{2}\) to show that \(\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\).

Another form of \(\int \sec x \, d x\) a. Verify the identity sec \(x=\frac{\cos x}{1-\sin ^{2} x}\) b. Use the identity in part (a) to verify that \(\int \sec x \, d x=\frac{1}{2} \ln \left|\frac{1+\sin x}{1-\sin x}\right|+C\)

Shortcut for Simpson's Rule Using the notation of the text, prove that \(S(2 n)=\frac{4 T(2 n)-T(n)}{3},\) for \(n \geq 1\)

Determine whether the following integrals converge or diverge. $$\int_{3}^{\infty} \frac{d x}{\ln x}(\text { Hint: } \ln x \leq x .)$$

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