Any tips on how to show that \int_0^\pi f(\sin x)dx=2\int_0^{\frac{\pi}{2}}f(\sin x)dx

untchick04tm

untchick04tm

Answered question

2022-01-04

Any tips on how to show that
0πf(sinx)dx=20π2f(sinx)dx

Answer & Explanation

alkaholikd9

alkaholikd9

Beginner2022-01-05Added 37 answers

Note that we have
0πf(sin(x))dx=0π2f(sin(x))dx+π2πf(sin(x))dx
=0π2f(sin(x))dx+π20f(sin(πt))(dt)
=0π2f(sin(x))dx+0π2f(sin(t))dt
=20π2f(sin(x))dx
Dabanka4v

Dabanka4v

Beginner2022-01-06Added 36 answers

Split the integral:
0πf(sin(x))=)π2f(sin(x))+π2πf(sin(x))
Shift one integral:
π2πf(sin(x))=0π2f(sin(x+π2))
Reverse:
0π2f(sin(x+π2))=0π2f(sin((π2x)+π2))=0π2f(sin(πx))
But since sin(x)=sin(πx) we can simplify:
0πf(sin(x))=20π2f(sin(x))
karton

karton

Expert2022-01-11Added 613 answers

Generalization:
02af(x)dx=0af(x)dx+a2af(x)dx
Now set 2ax=u and what happens if f(2ax)=f(x)

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