Can U-Substitution Prove This Trigonometric Integral Identity?

[Samuelb88]

help with u-substitution - prove...

Homework Statement

Some function f is continuous on [0,Pi]

Prove: \int_0^{\pi}\\xf(sin x)\,dx = \frac{\pi}{2}\right)\int_0^{\pi}\\f(sinx)dx

using the substitution u=\pi-x.

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 (\int_0^{\pi}\\xf(sin x)\,dx) to the one on the right and (2) vice versa.

1) If I am to use the u-sub: u=\Pi-x, then:

du=d(\Pi-x)=-dx <==> -du=dx

First, from the identity:

sin(\Pi-x)=sinx

I re-expressed the integral like such:

\int_0^{\pi}\\(\Pi-x)f(sin(\Pi-x))\,dx

Then substituted the quantity u into the integrand:

-\int_\Pi^{0}\\(u)f(sin(u))\,du

Therefore:

\int_0^{\Pi}\\(u)f(sin(u))\,du

Now here's where I get stuck, after much trial and error, ultimately leading to no where, I've found that \frac{d}{dx}\right)(\frac{F(x^2)}{2}\right)) = xf(x^2).

2) From the left side of the statement.

\frac{\pi}{2}\right)\int_0^{\pi}\\f(sinx)dx

When I start from the side, it appears as though the quantity u = \frac{2}{\Pi}\right)x and the differential du=\frac{2}{\Pi}\right)dx but then obviously I can't set u=Pi-x.

?

 

[jbunniii]

Homework Statement

Some function f is continuous on [0,Pi]

Prove: \int_0^{\pi}\\xf(sin x)\,dx = \frac{\pi}{2}\right)\int_0^{\pi}\\f(sinx)dx

using the substitution u=\pi-x.

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 (\int_0^{\pi}\\xf(sin x)\,dx) to the one on the right and (2) vice versa.

1) If I am to use the u-sub: u=\Pi-x, then:

du=d(\Pi-x)=-dx <==> -du=dx

First, from the identity:

sin(\Pi-x)=sinx

I re-expressed the integral like such:

\int_0^{\pi}\\(\Pi-x)f(sin(\Pi-x))\,dx

Then substituted the quantity u into the integrand:

-\int_\Pi^{0}\\(u)f(sin(u))\,du

Therefore:

\int_0^{\Pi}\\(u)f(sin(u))\,du

I'm not sure what you are doing here. This has taken you right back where you started.

If u = \pi - x then du = -dx as you indicated, so

\int_0^\pi x f(\sin x) dx = \int_0^\pi (\pi - u) f(\sin(\pi - u)) du =<br /> \int_0^\pi (\pi - u) f(\sin u) du

Note that there's nothing special about the letter u on the right hand side. You can just as well replace it with x. If you do that, the answer is actually right in front of your eyes, in a slightly tricky form.