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Using Trigonometric Substitution In Exercises \(5-8,\) find the indefinite integral using the substitution \(x=4 \sin \theta .\) $$ \int \frac{x^{3}}{\sqrt{16-x^{2}}} d x $$

Short Answer

Expert verified
The indefinite integral of the given function is \(-16\sqrt{1-(x/4)^2} + \frac{64}{12}*3/2\sqrt{1-(x/4)^2} - C\).

Step by step solution

01

Perform the Substitution

Substitute \(x=4 \sin \theta\) into the integral. This changes the integral into the variable \(\theta\), and requires also changing dx into terms of d\(\theta\). The derivative dx/d\(\theta\) = 4 cos\(\theta\) so dx = 4 cos\(\theta\) d\(\theta\).
02

Substitute into Integral

For \(x^3\), substitute \((4 \sin \theta)^3 = 64 \sin^3 \theta\). Then substitute \(\sqrt{16-x^2}\) with \(\sqrt{16-16\sin^2 \theta} = 4\cos\theta\). This simplifies the integral to:$$ \int \frac{64 \sin^3 \theta * 4\cos \theta}{4\cos\theta } d\theta $$.
03

Simplify the Integral

Simplify the integral by cancelling out \(4\cos\theta\) from the numerator and denominator, leading to the integral expression: $$ \int 64 \sin^3 \theta d\theta $$.
04

Integrate

Use the integral property of \(\sin^3 \theta = \frac{3}{4}\sin\theta - \frac{1}{4}\sin3\theta\) and calculate the integral which results in \(\frac{64}{4}(-\cos \theta + \frac{1}{3} \cos3\theta) = -16\cos \theta + \frac{64}{12} \cos3\theta\).
05

Resubstitute the original variable

Recall the original substitution \(x=4\sin\theta\). Hence, \(\sin\theta = \frac{x}{4}\) and \(\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - (\frac{x}{4})^2}\). Substitute these values back into the integral, resulting into \(-16\sqrt{1-(x/4)^2} + \frac{64}{12}*3/2\sqrt{1-(x/4)^2} - C\), where C is the constant of integration.

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Key Concepts

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

Indefinite Integrals
When we talk about indefinite integrals, we're discussing mathematical expressions that involve integration without specific limits or boundaries. In essence, an indefinite integral is the reverse process of differentiation. The result of this operation is a family of functions plus a constant denoted by 'C'. This is because differentiation annihilates constants, so when finding the original function from its derivative, we must include an arbitrary constant.

Indefinite integrals are represented by the integral sign followed by a function and dx, like this:
  • \( \int f(x) \, dx \)
The aim is to find a function F(x) such that when differentiated it gives back the original function f(x).

Keep in mind that indefinite integrals can become quite complex when dealing with non-linear functions or when additional variables and substitutions are involved.
Trigonometric Identities
Trigonometric identities are mathematical equations that involve trigonometric functions and are always true for all allowable values. They are particularly useful in simplifying expressions and solving integrals, especially when applying trigonometric substitution.

Some essential trigonometric identities used in calculus include:
  • Pythagorean Identity: \( \sin^2 \theta + \cos^2 \theta = 1 \)
  • Angle Sum and Difference Identities
  • Double Angle Identity
For this exercise, the Pythagorean Identity plays a crucial role. When using trigonometric substitution, such as \(x = 4 \sin \theta\), we often need to replace expressions like \(\sqrt{16 - x^2}\).

By recognizing the trigonometric form, \(\sqrt{16 - 16 \sin^2 \theta}\), you can simplify it to \(4 \cos \theta\) using the identity.
Integration Techniques
Integration techniques are methods used to solve complex integrals that cannot be solved by straightforward methods. Trigonometric substitution is one such technique and is very effective in handling integrals that include expressions under a square root involving quadratics.

For example, with the substitution \(x = 4 \sin \theta\), the integral \( \int \frac{x^3}{\sqrt{16-x^2}} \, dx \) can be transformed into a form involving trigonometric functions. The transformation simplifies the integral by eliminating the square root, leading to an expression that is easier to handle.
  • The substitution method often requires careful back-substitution to return to the original variable.
  • You need to find both \(x\) and \(dx\) in terms of \(\theta\) to execute the substitution.
This technique requires familiarity with trigonometric identities and inverse trigonometric functions to re-substitute the variable at the end of the process.
Calculus Problems
Calculus problems can vary widely in difficulty and scope. This exercise highlights how integration can be complicated by non-polynomial terms and radicals.

Such problems challenge students to apply multiple steps and techniques, including:
  • Identifying the correct substitution, like \(x = 4 \sin \theta\), to simplify the integral.
  • Recognizing and applying trigonometric identities to rewrite the integral without square roots.
  • Performing algebraic manipulation to simplify the integral to a more straightforward form before integration.
Finally, the back-substitution stage requires re-expressing the answer in terms of the original variable. Calculus problems involving indefinite integrals test a student's understanding of both integration fundamentals and strategic problem-solving abilities.

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