Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question
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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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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