If f is continuous on [0, π], use the substitution u = -x to show that -76 2 ["xf(s xf(sin(x)) dx = f(sin(x)) dx.

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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If f is continuous on [0, π], use the substitution u = - x to show that
[²x₁
/0
So we have,
π
[²x₁
xf(sin(x)) dx =
π
2 ["x
10
[T
10
---Select--- V When x = π, u = 0, and when x = 0, u =---Select---
Let u = π - x. Then du =
Use U-substitution to replace the values in the integral.
'0
["xf(sin(x)) dx = [ (ir - () (sin( [
L)
(π
Jπ
π
2
xf(sin(x)) dx = π
- ["(x - 1)r (sin( [
=
(π
f(sin(x)) dx.
Recall the identity that sin(π - u) = sin(u) and make the appropriate substitution and continue the proof.
["x²{(sin(x)) dx = ["(x - u)r (sin(
(π
)) du
-x [ "r(sin(us) on - ["ur(sin( [
=
du
= x (" (sin(x)) dx - (²xr(sin( [
x f(sin(x)) dx = π
--6²
7. STF (
f(sin(x)) dx
"π
= = ["r (sin( [
10
"π
- [² x r( sin ( [
(- du)
1)) a
dx
du
→
)) au
dx
介
])) dx =
Transcribed Image Text:If f is continuous on [0, π], use the substitution u = - x to show that [²x₁ /0 So we have, π [²x₁ xf(sin(x)) dx = π 2 ["x 10 [T 10 ---Select--- V When x = π, u = 0, and when x = 0, u =---Select--- Let u = π - x. Then du = Use U-substitution to replace the values in the integral. '0 ["xf(sin(x)) dx = [ (ir - () (sin( [ L) (π Jπ π 2 xf(sin(x)) dx = π - ["(x - 1)r (sin( [ = (π f(sin(x)) dx. Recall the identity that sin(π - u) = sin(u) and make the appropriate substitution and continue the proof. ["x²{(sin(x)) dx = ["(x - u)r (sin( (π )) du -x [ "r(sin(us) on - ["ur(sin( [ = du = x (" (sin(x)) dx - (²xr(sin( [ x f(sin(x)) dx = π --6² 7. STF ( f(sin(x)) dx "π = = ["r (sin( [ 10 "π - [² x r( sin ( [ (- du) 1)) a dx du → )) au dx 介 ])) dx =
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