To determine

To show:

∫0πxf(sin⁡x)dx=π2∫0πf(sin⁡x)dx

Answer

∫0πxf(sin⁡x)dx=π2∫0πf(sin⁡x)dx

Explanation

1) Concept:

i. The substitution rule: If u=g(x) is a differentiable function whose range is I, and f is continuous on I, then ∫f(gx)g'xdx=∫f(u)du. Here g(x) is substituted as u and  g’(x)dx =du

ii.

∫abf(x)dx=-∫baf(x)dx

iii.

∫abf(u)du=∫abf(x)dx

2) Given:

f  is continuous on [0, π]

3) Formula:

sin⁡π-x=sin⁡x

4) calculation:

Let

I= ∫0πxf(sin⁡x)dx

Here use the substitution method

Substitute π-x=u

That is the same as x=π-u

Differentiating with respect to x

dx=-du

The limits changes; the new limits of integration are calculated by substituting

At x=0, u=π-0=π and

At x=π, u=π-π=0

Therefore, the integral becomes

I= ∫0πxfsin⁡xdx

=∫π0(π-u)f(sin⁡(π-u))(-du)

Simplify and using sin⁡π-u=sin⁡u

=∫π0-(π-u)f(sin⁡u)du

Usingthe property of definite integral

=∫0π(π-u)f(sin⁡u)du

Now split the integral sign

=∫0ππf(sin⁡u)du-∫0πuf(sin⁡u)du

Using concept iii)

=∫0ππf(sin⁡x)dx-∫0πxf(sin⁡x)dx

Put I=∫0πxf(sin⁡x)dx thus we have

I=∫0ππf(sin⁡x)dx-I

Add I on both sides

I+I=∫0ππf(sin⁡x)dx-I+I

Simplify

2I=∫0ππf(sin⁡x)dx

Divide by 2 on both sides and simplify

I=π2∫0πf(sin⁡x)dx

Therefore,

∫0πxf(sin⁡x)dx=π2∫0πf(sin⁡x)dx

Conclusion:

Therefore,

∫0πxf(sin⁡x)dx=π2∫0πf(sin⁡x)dx