   Chapter 4.5, Problem 64E

Chapter
Section
Textbook Problem

# If f is continuous on [ 0 , π ] , use the substitution u = π − x to show that to show that ∫ 0 π x f ( sin x ) d x = π 2 ∫ 0 π f ( sin x ) d x

To determine

To show:

0πxf(sinx)dx=π20πf(sinx)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:

sinπ-x=sinx

4) calculation:

Let

I= 0πxf(sinx)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= 0πxfsinxdx

=π0(π-u)f(sin(π-u))(-du)

Simplify and using sinπ-u=sinu

=π0-(π-u)f(sinu)du

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