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Evaluate the integral. 17 sin?(x) cos (x) dx

Evaluate the integral. 17 sin?(x) cos (x) dx

Calculus: Early Transcendentals
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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# Evaluating the Integral Step-by-Step ### Step-by-Step Guide: We aim to evaluate the integral: \[ \int 17 \sin^2(x) \cos^3(x) \, dx \] ### Step 1 Since \( \int 17 \sin^2(x) \cos^3(x) \, dx \) has an odd power of \( \cos(x) \), we will convert all but one power to sines. We know that: \[ \cos^2(x) = 1 - \sin^2(x) \] ### Step 2 Making this substitution, the integral \[ \int 17 \sin^2(x) \cos^3(x) \, dx \] gives us: \[ \int 17 \sin^2(x) \cos(x) (1 - \sin^2(x)) \, dx = \int 17 \sin^2(x) \cos(x) \, dx - \int 17 \sin^4(x) \cos(x) \, dx \] ### Step 3 Since \( \cos(x) \) is the derivative of \( \sin(x) \), the integral: \[ \int 17 \sin^2(x) \cos(x) \, dx \] can be done by substituting \[ u = \sin(x) \] Thus, \[ du = \cos(x) \, dx \] ### Step 4 With the substitution \( u = \sin(x) \), we get: \[ \int 17 \sin^2(x) \cos(x) \, dx = 17 \int u^2 \, du \] which integrates to: \[ 17 \left( \frac{u^3}{3} \right) + C \] Substituting back in to get the answer in terms of \( \sin(x) \), we have: \[ \int 17 \sin^2(x) \cos(x) \, dx = 17 \left( \frac{\sin^3(x)}{3} \right) + C \] Therefore, the solution to the integral is: \[ 17 \left( \frac{\sin^3(x)}{3} - \frac{\sin^5(x)}{5} \right) + C \]
Transcribed Image Text:# Evaluating the Integral Step-by-Step ### Step-by-Step Guide: We aim to evaluate the integral: \[ \int 17 \sin^2(x) \cos^3(x) \, dx \] ### Step 1 Since \( \int 17 \sin^2(x) \cos^3(x) \, dx \) has an odd power of \( \cos(x) \), we will convert all but one power to sines. We know that: \[ \cos^2(x) = 1 - \sin^2(x) \] ### Step 2 Making this substitution, the integral \[ \int 17 \sin^2(x) \cos^3(x) \, dx \] gives us: \[ \int 17 \sin^2(x) \cos(x) (1 - \sin^2(x)) \, dx = \int 17 \sin^2(x) \cos(x) \, dx - \int 17 \sin^4(x) \cos(x) \, dx \] ### Step 3 Since \( \cos(x) \) is the derivative of \( \sin(x) \), the integral: \[ \int 17 \sin^2(x) \cos(x) \, dx \] can be done by substituting \[ u = \sin(x) \] Thus, \[ du = \cos(x) \, dx \] ### Step 4 With the substitution \( u = \sin(x) \), we get: \[ \int 17 \sin^2(x) \cos(x) \, dx = 17 \int u^2 \, du \] which integrates to: \[ 17 \left( \frac{u^3}{3} \right) + C \] Substituting back in to get the answer in terms of \( \sin(x) \), we have: \[ \int 17 \sin^2(x) \cos(x) \, dx = 17 \left( \frac{\sin^3(x)}{3} \right) + C \] Therefore, the solution to the integral is: \[ 17 \left( \frac{\sin^3(x)}{3} - \frac{\sin^5(x)}{5} \right) + C \]
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