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Samuelb88
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help with u-substitution - prove...
Some function f is continuous on [0,Pi]
Prove: [tex]\int_0^{\pi}\\xf(sin x)\,dx = \frac{\pi}{2}\right)\int_0^{\pi}\\f(sinx)dx[/tex]
using the substitution [tex]u=\pi-x[/tex].
Identities, integral properties.
I've tried this two ways, (1) trying to transform the integral on the left ([tex]\int_0^{\pi}\\xf(sin x)\,dx[/tex]) to the one on the right and (2) vice versa.
1) If I am to use the u-sub: [tex]u=\Pi-x[/tex], then:
[tex]du=d(\Pi-x)=-dx[/tex] <==> [tex]-du=dx[/tex]
First, from the identity:
[tex]sin(\Pi-x)=sinx[/tex]
I re-expressed the integral like such:
[tex]\int_0^{\pi}\\(\Pi-x)f(sin(\Pi-x))\,dx[/tex]
Then substituted the quantity u into the integrand:
[tex]-\int_\Pi^{0}\\(u)f(sin(u))\,du[/tex]
Therefore:
[tex]\int_0^{\Pi}\\(u)f(sin(u))\,du[/tex]
Now here's where I get stuck, after much trial and error, ultimately leading to no where, I've found that [tex]\frac{d}{dx}\right)(\frac{F(x^2)}{2}\right)) = xf(x^2)[/tex].
2) From the left side of the statement.
[tex]\frac{\pi}{2}\right)\int_0^{\pi}\\f(sinx)dx[/tex]
When I start from the side, it appears as though the quantity [tex]u = \frac{2}{\Pi}\right)x[/tex] and the differential [tex]du=\frac{2}{\Pi}\right)dx[/tex] but then obviously I can't set u=Pi-x.
?
Homework Statement
Some function f is continuous on [0,Pi]
Prove: [tex]\int_0^{\pi}\\xf(sin x)\,dx = \frac{\pi}{2}\right)\int_0^{\pi}\\f(sinx)dx[/tex]
using the substitution [tex]u=\pi-x[/tex].
Homework Equations
Identities, integral properties.
The Attempt at a Solution
I've tried this two ways, (1) trying to transform the integral on the left ([tex]\int_0^{\pi}\\xf(sin x)\,dx[/tex]) to the one on the right and (2) vice versa.
1) If I am to use the u-sub: [tex]u=\Pi-x[/tex], then:
[tex]du=d(\Pi-x)=-dx[/tex] <==> [tex]-du=dx[/tex]
First, from the identity:
[tex]sin(\Pi-x)=sinx[/tex]
I re-expressed the integral like such:
[tex]\int_0^{\pi}\\(\Pi-x)f(sin(\Pi-x))\,dx[/tex]
Then substituted the quantity u into the integrand:
[tex]-\int_\Pi^{0}\\(u)f(sin(u))\,du[/tex]
Therefore:
[tex]\int_0^{\Pi}\\(u)f(sin(u))\,du[/tex]
Now here's where I get stuck, after much trial and error, ultimately leading to no where, I've found that [tex]\frac{d}{dx}\right)(\frac{F(x^2)}{2}\right)) = xf(x^2)[/tex].
2) From the left side of the statement.
[tex]\frac{\pi}{2}\right)\int_0^{\pi}\\f(sinx)dx[/tex]
When I start from the side, it appears as though the quantity [tex]u = \frac{2}{\Pi}\right)x[/tex] and the differential [tex]du=\frac{2}{\Pi}\right)dx[/tex] but then obviously I can't set u=Pi-x.
?
Last edited: