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.