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Trigonometric substitutions Evaluate the following integrals using trigonometric substitution. $$\int \frac{x^{2}}{\sqrt{16-x^{2}}} d x$$

Short Answer

Expert verified
Answer: \(\frac{x}{8} + C\)

Step by step solution

01

Identify the trigonometric substitution

Since we have \(\sqrt{16 - x^2}\), we can use the substitution $$x = 4\sin{θ}$$. Now we need to find the derivative of this substitution: $$\frac{d x}{d θ} = 4\cos{θ}$$
02

Rewrite the integral in terms of $$θ$$

Replace $$x$$ with $$4\sin{θ}$$: $$\int \frac{(4\sin{θ})^{2}}{\sqrt{16-(4\sin{θ})^{2}}} d x$$ Now we need to replace $$d x$$ with $$d θ$$. To do this, multiply both sides of the equation $$\frac{d x}{d θ} = 4\cos{θ}$$ by $$d θ$$ to get $$d x = 4\cos{θ} d θ$$. Replace $$d x$$ in the integral to get: $$\int \frac{(4\sin{θ})^{2}}{\sqrt{16-(4\sin{θ})^{2}}} (4 \cos{θ} d θ)$$ Now the integral is fully in terms of $$θ$$.
03

Simplify and evaluate the integral

Simplify the integral expression: $$\int \frac{16\sin^2{θ}}{\sqrt{16-16\sin^2{θ}}}(4 \cos{θ} d θ)$$ Divide numerator and denominator by 16 and notice that 1-sin²{θ} = cos²{θ}: $$\int \frac{\sin^2{θ}}{\sqrt{\cos^2{θ}}} (4 \cos{θ} d θ)$$ Now we can cancel the square root, because $$\sqrt{\cos^2{θ}} = \cos{θ}$$: $$\int \sin^2{θ} (4 \cos{θ} d θ)$$ Now perform the integration: $$4 \int \sin^2{θ}\cos{θ} dθ$$ Use the double angle identity to simplify further: \(\sin^2{θ} = \frac{1-\cos{2θ}}{2}\) $$\Rightarrow 2\int \left(\frac{1-\cos{2θ}}{2}\right)\cos{θ} dθ = \int(1-\cos{2θ})\cos{θ}dθ$$ Now, evaluate the integral: $$\int (1-\cos{2θ})\cos{θ}dθ = \int\cos{θ}dθ - \int\cos{2θ}\cos{θ}dθ = \sin{θ} - \frac{1}{2}\sin{θ}$$
04

Convert back to $$x$$

Now we need to rewrite the result in terms of $$x$$ using the substitution $$x=4\sin{θ}$$. We know that \(\sin{θ} = \frac{x}{4}\). So, our answer is: $$\sin{θ} - \frac{1}{2}\sin{θ} = \frac{x}{4} - \frac{1}{2}\frac{x}{4} = \frac{1}{2}\frac{x}{4} = \frac{x}{8} + C$$ Finally, the evaluated integral is: $$\int \frac{x^{2}}{\sqrt{16-x^{2}}} dx = \frac{x}{8} + C$$

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Key Concepts

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

Integration Techniques
Integration techniques in calculus are essential tools that help us solve complex integrals. One such technique is trigonometric substitution. Trigonometric substitution is often employed when an integral contains a square root of the form \( \sqrt{a^2 - x^2} \), \( \sqrt{a^2 + x^2} \), or \( \sqrt{x^2 - a^2} \). By substituting one of the trigonometric identities, such as \( x = a\sin{\theta} \), \( x = a\tan{\theta} \), or \( x = a\sec{\theta} \), the integral becomes easier to solve, as it converts into trigonometric functions, which are often more manageable.

In the example exercise, we used \( x = 4\sin{\theta} \) to take advantage of the identity \( \sin^2{\theta} + \cos^2{\theta} = 1 \), transforming the integral into a trigonometric one. This substitution helps manage the complexity inside the integral by converting the function of \( x \) into a function of \( \theta \).

Understanding these techniques is vital, as they appear in various mathematical and physical applications. By practicing trigonometric substitution, you will become more proficient in solving complex integrals that incorporate square-root expressions.
Trigonometric Identities
Trigonometric identities play a crucial role in simplifying and solving integrals. They transform complex trigonometric expressions into simpler ones. Some basic identities include \( \sin^2{\theta} + \cos^2{\theta} = 1 \) and the double-angle identities like \( \sin{2\theta} = 2\sin{\theta}\cos{\theta} \).In the given exercise, we applied the identity \( \sin^2{\theta} = \frac{1 - \cos{2\theta}}{2} \) to simplify the integral further. This identity lets us express \( \sin^2{\theta} \) in terms of \( \cos{2\theta} \), making the integral easier to handle.

When performing trigonometric substitutions, recognize the identities you can use. They streamline calculations and reduce errors by turning complex expressions into simpler forms. Knowing these identities expands your ability to work with integrals and extends your skill set for tackling diverse mathematical problems.

As you become more familiar with these identities, your problem-solving becomes more efficient and intuitive, making calculus integrals much less daunting.
Calculus Integrals
Integrals are a fundamental concept in calculus, capturing the accumulation of quantities and determining areas under curves. The process of solving an integral is called integration, which can often be as straightforward as finding antiderivatives or as complex as employing special techniques.

Trigonometric substitution is a specific strategy for challenging integrals, particularly those involving radicals. By converting the original variables into trigonometric forms, integrals become more approachable and more systematic to solve.

In the original exercise, we tackled an integral with \( \sqrt{16 - x^2} \), which is tricky because of the square root. Converting \( x \) into a function of \( \theta \) using \( x = 4\sin{\theta} \) simplifies the integral. Meanwhile, the conversion lets us apply known trigonometric identities to further transform and integrate the expression.

Understanding these transformation techniques not only equips you to solve integrals but also enhances your comprehension of areas and volumes, motion and growth rates, and several other crucial concepts in both mathematics and applied sciences.

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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 \(16 .\)

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