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Magazine Chapter 11: Problem 24
Evaluate the definite integral. $$ \int_{0}^{1} x \sin ^{-1} x d x $$
Short Answer
Expert verified
\( \frac{\pi}{8} \)
Step by step solution
01
Identify the integral and its limits
We need to evaluate the definite integral \(\int_{0}^{1} x \, \text{sin}^{-1}(x) \, dx\). The limits of integration are from 0 to 1.
02
Use integration by parts formula
To solve \(\int_{0}^{1} x \, \text{sin}^{-1}(x) \, dx\), use integration by parts: \(\int u \, dv = uv - \int v \, du\), where we choose \(u = \text{sin}^{-1}(x)\) and \(dv = x \, dx\).
03
Compute derivatives and integrals
Calculate \(du\) and \(v\): \(u = \text{sin}^{-1}(x) \rightarrow du = \frac{1}{\sqrt{1-x^2}} \, dx\), and \(dv = x \, dx \rightarrow v = \frac{x^2}{2}\).
04
Apply the integration by parts formula
Substitute \(u\), \(du\), \(v\), and \(dv\) into the integration by parts formula: \(\int_{0}^{1} x \, \text{sin}^{-1}(x) \, dx = \left. \frac{x^2}{2} \, \text{sin}^{-1}(x) \right|_{0}^{1} - \int_{0}^{1} \frac{x^2}{2} \, \frac{1}{\sqrt{1-x^2}} \, dx\).
05
Evaluate the boundary term
Evaluate the boundary term \(\left. \frac{x^2}{2} \, \text{sin}^{-1}(x) \right|_{0}^{1} = \left( \frac{1}{2} \, \text{sin}^{-1}(1) \right) - \left( \frac{0}{2} \, \text{sin}^{-1}(0) \right) = \frac{\pi}{4}\).
06
Simplify the remaining integral
Simplify the remaining integral \(\int_{0}^{1} \frac{x^2}{2} \, \frac{1}{\sqrt{1-x^2}} \, dx\). This simplifies to \(\frac{1}{2} \int_{0}^{1} \frac{x^2}{\sqrt{1-x^2}} \, dx\).
07
Substitute \(x = \sin(\theta)\)
Make the substitution \(x = \sin(\theta)\). Then \(dx = \cos(\theta) \, d\theta\) and the limits change from \(x = 0\) to \(\theta = 0\), and \(x = 1\) to \(\theta = \frac{\pi}{2}\).
08
Evaluate the integral using the substitution
Rewrite the integral: \(\frac{1}{2} \int_{0}^{1} \frac{x^2}{\sqrt{1-x^2}} \, dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\sin^2(\theta)}{\cos(\theta)} \, \cos(\theta) \, d\theta = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin^2(\theta) \, d\theta\).
09
Utilize trigonometric identity
Utilize the identity \(\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}\): \(\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin^2(\theta) \, d\theta = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos(2\theta)}{2} \, d\theta = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} (1 - \cos(2\theta)) \, d\theta\).
10
Integrate and solve the definite integral
Split the integral and solve: \(\frac{1}{4} \int_{0}^{\frac{\pi}{2}} 1 \, d\theta - \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \cos(2\theta) \, d\theta = \frac{1}{4} \left. \theta \right|_{0}^{\frac{\pi}{2}} - \frac{1}{8} \left. \sin(2\theta) \right|_{0}^{\frac{\pi}{2}} = \frac{1}{4} \cdot \frac{\pi}{2} - 0 = \frac{\pi}{8}\).
11
Combine results
Combine results from Step 5 and Step 10: \(\frac{\pi}{4} - \frac{\pi}{8} = \frac{2\pi}{8} - \frac{\pi}{8} = \frac{\pi}{8}\).
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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 technique used to integrate the product of two functions. The formula is given by:
\int u \, dv = uv - \int v \, du.
In our exercise, we apply it to solve \int_{0}^{1} x \, \text{sin}^{-1}(x) \, dx. Here's how to proceed:
\int_{0}^{1} x \, \text{sin}^{-1}(x) \, dx = \left. \frac{x^2}{2} \, \text{sin}^{-1}(x) \right|_{0}^{1} - \int_{0}^{1} \frac{x^2}{2} \, \frac{1}{\sqrt{1 - x^2}} \, dx.
Evaluate the boundary terms and simplify the remaining integral for the next steps, setting the stage for substitution or further simplification.
\int u \, dv = uv - \int v \, du.
In our exercise, we apply it to solve \int_{0}^{1} x \, \text{sin}^{-1}(x) \, dx. Here's how to proceed:
- Choose parts to differentiate and integrate: let \(u = \text{sin}^{-1}(x)\) and \(dv = x \, dx\).
- Differentiate \(u\): \(du = \frac{1}{\sqrt{1 - x^2}} \, dx\).
- Integrate \(dv\): \(v = \frac{x^2}{2}\).
\int_{0}^{1} x \, \text{sin}^{-1}(x) \, dx = \left. \frac{x^2}{2} \, \text{sin}^{-1}(x) \right|_{0}^{1} - \int_{0}^{1} \frac{x^2}{2} \, \frac{1}{\sqrt{1 - x^2}} \, dx.
Evaluate the boundary terms and simplify the remaining integral for the next steps, setting the stage for substitution or further simplification.
Inverse Trigonometric Functions
Inverse trigonometric functions鈥攍ike \(\sin^{-1}(x)\)鈥攑lay a crucial role in calculus. Ensure you understand their properties and derivatives:
- The inverse sine function, \(\sin^{-1}(x)\), outputs angles whose sine is \(x\).
- Its derivative is \(\frac{d}{dx} (\sin^{-1}(x)) = \frac{1}{\sqrt{1 - x^2}}\).
Trigonometric Substitution
Trigonometric substitution helps simplify integrals involving radicals. When you see expressions like \(\sqrt{1 - x^2}\), try substituting \(x = \sin(\theta)\). Here's how:
\(\frac{1}{2} \int_{0}^{1} \frac{x^2}{\sqrt{1-x^2}} \, dx\) becomes \(\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\sin^2(\theta)}{\cos(\theta)} \, \cos(\theta) \, d\theta = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin^2(\theta) \, d\theta\). This method simplifies the integral using known trigonometric identities, enabling more straightforward evaluation.
- If \(x = \sin(\theta)\), then \(dx = \cos(\theta) \, d\theta\).
- Also, \(\sqrt{1 - x^2} = \sqrt{1 - \sin^2(\theta)} = \cos(\theta)\).
\(\frac{1}{2} \int_{0}^{1} \frac{x^2}{\sqrt{1-x^2}} \, dx\) becomes \(\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\sin^2(\theta)}{\cos(\theta)} \, \cos(\theta) \, d\theta = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin^2(\theta) \, d\theta\). This method simplifies the integral using known trigonometric identities, enabling more straightforward evaluation.
