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Magazine Chapter 8: Problem 50
Evaluate the integrals. Some integrals do not require integration by parts. $$ \int_{0}^{1 / \sqrt{2}} 2 x \sin ^{-1}\left(x^{2}\right) d x $$
Short Answer
Expert verified
Using integration by parts, the integral evaluates to \( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \).
Step by step solution
01
Recognize the Type of Integral
The integral given is \( \int_{0}^{1/\sqrt{2}} 2x \sin^{-1}(x^2) \, dx \). This does not straightforwardly match the basic integral forms and seems suitable for integration by parts due to the composition \( \sin^{-1}(x^2) \).
02
Choose u and dv for Integration by Parts
Choose \( u = \sin^{-1}(x^2) \) and \( dv = 2x \, dx \). This is because the derivative of \( \sin^{-1}(x^2) \) will simplify the integral.
03
Differentiate u and Integrate dv
Differentiate \( u \): \( du = \frac{1}{\sqrt{1-(x^2)^2}} \cdot 2x \, dx \). Integrate \( dv \): \( v = x^2 \).
04
Apply Integration by Parts Formula
The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Substitute the values: \(uv = x^2 \sin^{-1}(x^2) \) and compute the remaining integral \( \int x^2 \frac{2x}{\sqrt{1-x^4}} \, dx \).
05
Simplify the Remaining Integral
The integral becomes \( \int \frac{2x^3}{\sqrt{1-x^4}} \, dx \). Use substitution: let \( z = x^2 \rightarrow dz = 2x \, dx \), simplifying the integral to \( \int \frac{z}{\sqrt{1-z^2}} \, dz \).
06
Evaluate the Remaining Integral
The simplified form \( \int \frac{z}{\sqrt{1-z^2}} \, dz \) is amenable to direct integration, yielding \( -\sqrt{1-z^2} \) as the integral result back in terms of \( x^4 \).
07
Apply Limits and Compute Final Integral
Substitute back to \( x \) terms where needed, then evaluate limits from 0 to \( 1/\sqrt{2} \) for both the derived expressions from parts and this final term. Calculate each contribution and sum them.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
Definite integrals are a powerful mathematical tool used to calculate the net area under a curve over a specific interval. Unlike indefinite integrals, they have limits of integration, which define the start and end points of the interval. Here's how it works:
To evaluate a definite integral between limits, follow these steps:
To evaluate a definite integral between limits, follow these steps:
- Perform the integration as you would for an indefinite integral.
- Apply the limits of integration by substituting the upper limit into the integrated function.
- Subtract the result of substituting the lower limit to find the final value.
Inverse Trigonometric Functions
Inverse trigonometric functions, such as \( \sin^{-1}(x) \), help us find angles when the values of trigonometric functions are known. In the original problem, \( \sin^{-1}(x^2) \) indicates that we should look for an angle whose sine is \( x^2 \).
These functions have unique properties:
These functions have unique properties:
- They are the inverses of the basic trigonometric functions like sine, cosine, and tangent.
- The range of \( \sin^{-1}(x) \) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\), making it crucial in calculating accurate angle measurements.
- They can complicate the integration process, often requiring methods like integration by parts to resolve.
U-Substitution
U-substitution is a technique used to simplify integration by changing variables, which makes an integral more manageable. Here's how it's applied:
When you encounter complex integrals, follow these steps:
When you encounter complex integrals, follow these steps:
- Identify a portion of the integral that, when substituted, simplifies the expression.
- Define \( u \) as this portion; differentiate to find \( du \).
- Rewrite the entire integral in terms of \( u \) and \( du \), simplifying where necessary.
Limits of Integration
The limits of integration, in our integral problem, were given as \( 0 \) and \( 1/\sqrt{2} \). They are crucial because they define the interval over which the definite integral is calculated.
When applying limits of integration:
When applying limits of integration:
- Evaluate the antiderivative function first.
- Replace the variable in your antiderivative with the upper limit, then the lower limit.
- Subtract the lower limit evaluation from the upper limit evaluation.
