As per the trigonometric substitution x = 4 sin 0, the following triangle is obtained. sin 0 = X x² 16x² (16-x The indefinite integral is X cos = V 16- X 4 Rewrite the equation in terms of x. 1- 8 1/2 J dx = 8 arcsin x² 16x2 dx = X XV 16- X 4 + C
As per the trigonometric substitution x = 4 sin 0, the following triangle is obtained. sin 0 = X x² 16x² (16-x The indefinite integral is X cos = V 16- X 4 Rewrite the equation in terms of x. 1- 8 1/2 J dx = 8 arcsin x² 16x2 dx = X XV 16- X 4 + C
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Trigonometric Substitution in Integrals
**Step 4**
As per the trigonometric substitution \( x = 4 \sin \theta \), the following triangle is obtained.
#### Diagram Description:
- A right triangle is presented with:
- One side labeled as \( x \).
- The hypotenuse labeled as \( 4 \).
- One angle labeled as \( \theta \).
- The other side labeled as \( \sqrt{16 - x^2} \).
#### Trigonometric Identities:
From the triangle, we have:
\[ \sin \theta = \frac{x}{4} \]
\[ \cos \theta = \frac{\sqrt{16 - x^2}}{4} \]
#### Equation Rewrite:
Rewrite the equation in terms of \( x \):
\[ \int \frac{x^2}{\sqrt{16 - x^2}} \, dx = 8 \left( \arcsin \frac{x}{4} - \frac{x}{4} \frac{\sqrt{16 - x^2}}{4} \right) + C \]
#### Indefinite Integral:
The indefinite integral is:
\[ \int \frac{x^2}{\sqrt{16 - x^2}} \, dx = \boxed{\text{Solution}} \]
---
Please fill in the solution in the boxed area after completing the integration process.
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Transcribed Image Text:### Trigonometric Substitution in Integrals
**Step 4**
As per the trigonometric substitution \( x = 4 \sin \theta \), the following triangle is obtained.
#### Diagram Description:
- A right triangle is presented with:
- One side labeled as \( x \).
- The hypotenuse labeled as \( 4 \).
- One angle labeled as \( \theta \).
- The other side labeled as \( \sqrt{16 - x^2} \).
#### Trigonometric Identities:
From the triangle, we have:
\[ \sin \theta = \frac{x}{4} \]
\[ \cos \theta = \frac{\sqrt{16 - x^2}}{4} \]
#### Equation Rewrite:
Rewrite the equation in terms of \( x \):
\[ \int \frac{x^2}{\sqrt{16 - x^2}} \, dx = 8 \left( \arcsin \frac{x}{4} - \frac{x}{4} \frac{\sqrt{16 - x^2}}{4} \right) + C \]
#### Indefinite Integral:
The indefinite integral is:
\[ \int \frac{x^2}{\sqrt{16 - x^2}} \, dx = \boxed{\text{Solution}} \]
---
Please fill in the solution in the boxed area after completing the integration process.
**Navigation Buttons:**
- **Submit**: To submit your answers.
- **Skip (you cannot come back)**: To skip this step without the possibility to return.
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