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Magazine Chapter 28: Problem 16
Integrate each of the functions. $$\int \frac{\sin ^{-1} 4 x d x}{\sqrt{1-16 x^{2}}}$$
Short Answer
Expert verified
The integral evaluates to \( x \sin^{-1}(4x) + 8x^2 + C \).
Step by step solution
01
Identify the Structure
The function to integrate is \( \int \frac{\sin^{-1}(4x)}{\sqrt{1 - 16x^2}} \, dx \). This resembles the derivative of the arcsine function. Recall that the derivative of \( \sin^{-1}(u) \) is \( \frac{1}{\sqrt{1-u^2}} \), where \( u = 4x \).
02
Set the Substitution
To simplify the integral, let's set \( u = 4x \). Then, the differential \( du = 4 \, dx \), or rearranged, \( dx = \frac{1}{4} du \). This substitution will transform the integral into simpler parts.
03
Substitute and Simplify
Substituting \( u = 4x \) and \( dx = \frac{1}{4} du \) into the integral, we have:\[\int \frac{\sin^{-1}(u)}{\sqrt{1-u^2}} \cdot \frac{1}{4} \, du = \frac{1}{4} \int \sin^{-1}(u) \, du\]
04
Use Integration by Parts
Use the integration by parts formula, \( \int v \, du = uv - \int u \, dv \). Choose \( v = \sin^{-1}(u) \) and \( du = du \) to solve for the integral. Then, differentiate to find \( dv = \frac{1}{\sqrt{1-u^2}} \, du \) and integrate \( \int du = u \). Apply these in the integration by parts formula.
05
Calculate Integration by Parts
Applying integration by parts:\[\frac{1}{4} \left[ u \sin^{-1}(u) \bigg| - \int u \cdot \frac{1}{\sqrt{1-u^2}} \, du \right]\]The integral \( \int u \cdot \frac{1}{\sqrt{1-u^2}} \, du \) simplifies with substitution \( u = \sin \theta \) to calculate further, eventually integrating to \(-\frac{1}{2}u^2 \).
06
Final Integration Result
Completing the calculations from the integration by parts result, we have:\[\frac{1}{4} \left[ u \sin^{-1}(u) + \frac{1}{2} \cdot u^2 + C \right]\]Replace back \( u = 4x \) leading to the final integral result:
07
Substitute Back
Substitute \( u = 4x \) back into the expression obtained after integration by parts.\[\frac{1}{4} \left[ 4x \sin^{-1}(4x) + \frac{1}{2}(4x)^2 + C \right]\]Simplify this to get the final integrated result. The integral evaluates to:\[x \sin^{-1}(4x) + 8x^2 + C\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by parts is a calculus technique used to integrate complex functions by breaking them down into simpler parts. It is represented by the formula:\[ \int u \, dv = uv - \int v \, du \]This method is particularly helpful when dealing with products of functions or cases where one function simplifies upon subdivision.
- Identify the parts: Choose two components, typically denoted as \( u \) and \( dv \).
- Differentiate \( u \) to get \( du \) and integrate \( dv \) to find \( v \).
- Substitute into the integration by parts formula.
- Solve the remaining integral \( \int v \, du \).
Substitution Method
The substitution method, also known as \( u \)-substitution, is an essential technique to simplify complex integrals. The main idea is to transform the integral into a simpler form by introducing a new variable \( u \).
- Identify a part of the integral to substitute, often the function inside another function, such as \( u = 4x \) here.
- Find \( du \) and rearrange to express \( dx \) in terms of \( du \).
- Substitute \( u \) and \( du \) into the integral, converting it to an easier form.
- Solve the integral with respect to \( u \).
- Finally, substitute back the original variable in place of \( u \).
Inverse Trigonometric Functions
Inverse trigonometric functions, such as \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \), are functions that reverse the usual trigonometric operations. They have specific derivatives that are often encountered in calculus problems:
- The derivative of \( \sin^{-1}(u) \) is \( \frac{1}{\sqrt{1-u^2}} \).
- Inverse functions are useful in solving integrals involving expressions like \( \sqrt{1-x^2} \).
