   Chapter 5.5, Problem 92E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# If f is continuous on [0, π], use the substitution u = π − x to show that ∫ 0 π x f ( sin x )   d x = π 2 ∫ 0 π f ( sin x )   d x

To determine

To prove: The integral function, 0πxf(sinx)dx=π20πf(sinx)dx using substitution.

Explanation

Given:

The integral function of left hand side (LHS) is 0πxf(sinx)dx.

The region lies between x=0 and x=π.

The substitution is u=πx

Calculation:

Consider u=πx

u=πxx=πu (1)

Differentiate both sides of the Equation (1).

dx=du

Calculate the lower limit value of u using Equation (1).

Substitute 0 for x in Equation (1).

u=π0=π

Calculate the upper limit value of u using Equation (1).

Substitute π for x in Equation (1).

u=ππ=0

The integral function is,

0πxf(sinx)dx (2)

Apply lower and upper limits for u in Equation (2).

Substitute (πu) for x and (du) for dx in Equation (2).

0πxf(sinx)dx=π0[(πu)f[sin(πu)]](du)=π0[(πu)f[sin(πu)]]du=π0[(πu)f[sinπsinu]]du=π0[(πu)f[0sinu]]du=0πf(πu)(sinu)du=0π[

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Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th

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