To solve the given integral using the substitution method, we can substitute u = π - x and find the new limits of integration. Then, we perform the substitution by replacing x with u and dx with -du. Finally, we can rewrite the integral using the fact that ∫0π f(x) dx = ∫0π f(t) dt:
∫0π x f(sin x) dx = π ∫0π (π - u) f(sin u) du = (π/2) ∫0π f(sin u) du
Step-by-step explanation:
To solve the given integral, we can use the substitution u = π - x. Let's first find the new limits of integration. When x = 0, u = π - 0 = π. And when x = π, u = π - π = 0. So the new limits of integration are u = 0 to u = π.
Next, let's find the derivative du/dx. Since u = π - x, we can rearrange it as x = π - u. Taking the derivative of both sides with respect to x, du/dx = -1.
Now we can perform the substitution by replacing x with u and dx with -du. The integral becomes:
∫0π x f(sin x) dx = ∫π0 (π - u) f(sin(π - u)) (-du)
Using the properties of integrals, we can pull out the negative sign and factor π out of the integral:
∫0π x f(sin x) dx = -π ∫0π (u - π) f(sin(π - u)) du
Simplifying further, we get:
∫0π x f(sin x) dx = -π ∫0π (u - π) f(sin u) du
Now we can use the property of integrals that states ∫ab f(x) dx = ∫ab f(t) dt. Applying this property to our integral, we have:
∫0π x f(sin x) dx = -π ∫0π (u - π) f(sin u) du = π ∫0π (π - u) f(sin u) du
Finally, we can rewrite the integral using the fact that ∫0π f(x) dx = ∫0π f(t) dt:
∫0π x f(sin x) dx = π ∫0π (π - u) f(sin u) du = (π/2) ∫0π f(sin u) du