Show that `int_0^pifx(sinx)dx=pi/2int_0^pif(sinx)dxdot`

Explanation

To show that

\[ \int_0^\pi f(x) \sin x \, dx = \frac{\pi}{2} \int_0^\pi f(\sin x) \, dx, \]

we will follow these steps:

### Step 1: Define the integral
Let

\[ I = \int_0^\pi f(x) \sin x \, dx. \]

### Step 2: Use the property of definite integrals
We can use the property of definite integrals which states that

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx. \]

In our case, we take \( a = 0 \) and \( b = \pi \). Thus, we have:

\[ I = \int_0^\pi f(\pi - x) \sin(\pi - x) \, dx. \]

### Step 3: Simplify the integral
Using the identity \( \sin(\pi - x) = \sin x \), we can rewrite the integral:

\[ I = \int_0^\pi f(\pi - x) \sin x \, dx. \]

### Step 4: Add the two integrals
Now, we add the two expressions for \( I \):

\[ 2I = \int_0^\pi f(x) \sin x \, dx + \int_0^\pi f(\pi - x) \sin x \, dx. \]

### Step 5: Combine the integrals
This can be combined as:

\[ 2I = \int_0^\pi \left( f(x) + f(\pi - x) \right) \sin x \, dx. \]

### Step 6: Change of variable
Next, we will change the variable in the second integral. Let \( u = \sin x \). The limits of integration remain the same, and we have:

\[ du = \cos x \, dx. \]

When \( x = 0 \), \( u = 0 \) and when \( x = \pi \), \( u = 0 \). However, we need to consider the symmetry of the sine function. The integral can be expressed as:

\[ \int_0^\pi f(\sin x) \, dx = \int_0^1 f(u) \frac{1}{\sqrt{1-u^2}} \, du. \]

### Step 7: Final expression
Thus, we can express \( 2I \) in terms of the integral of \( f(\sin x) \):

\[ 2I = \pi \int_0^1 f(u) \, du. \]

### Step 8: Solve for \( I \)
Dividing both sides by 2, we get:

\[ I = \frac{\pi}{2} \int_0^\pi f(\sin x) \, dx. \]

### Conclusion
Thus, we have shown that:

\[ \int_0^\pi f(x) \sin x \, dx = \frac{\pi}{2} \int_0^\pi f(\sin x) \, dx. \]