Key to "Any tips on how to show that ∫0πf(sin⁡x)dx=2∫0π2f(sin⁡x)dx"

untchick04tm

Answered question

2022-01-04

Any tips on how to show that

\int_{0}^{π} f (\sin x) d x = 2 \int_{0}^{\frac{π}{2}} f (\sin x) d x

Answer & Explanation

alkaholikd9

Beginner2022-01-05Added 37 answers

Note that we have

\int_{0}^{π} f (\sin (x)) d x = \int_{0}^{\frac{π}{2}} f (\sin (x)) d x + \int_{\frac{π}{2}}^{π} f (\sin (x)) d x

= \int_{0}^{\frac{π}{2}} f (\sin (x)) d x + \int_{\frac{π}{2}}^{0} f (\sin (π - t)) (- d t)

= \int_{0}^{\frac{π}{2}} f (\sin (x)) d x + \int_{0}^{\frac{π}{2}} f (\sin (t)) d t

= 2 \int_{0}^{\frac{π}{2}} f (\sin (x)) d x

Dabanka4v

Beginner2022-01-06Added 36 answers

Split the integral:

\int_{0}^{π} f (\sin (x)) = \int_{)}^{\frac{π}{2}} f (\sin (x)) + \int_{\frac{π}{2}}^{π} f (\sin (x))

Shift one integral:

\int_{\frac{π}{2}}^{π} f (\sin (x)) = \int_{0}^{\frac{π}{2}} f (\sin (x + \frac{π}{2}))

Reverse:

\int_{0}^{\frac{π}{2}} f (\sin (x + \frac{π}{2})) = \int_{0}^{\frac{π}{2}} f (\sin ((\frac{π}{2} - x) + \frac{π}{2})) = \int_{0}^{\frac{π}{2}} f (\sin (π - x))

But since

\sin (x) = \sin (π - x)

we can simplify:

\int_{0}^{π} f (\sin (x)) = 2 \int_{0}^{\frac{π}{2}} f (\sin (x))

karton

Expert2022-01-11Added 613 answers

Generalization:
$\int_{0}^{2 a} f (x) d x = \int_{0}^{a} f (x) d x + \int_{a}^{2 a} f (x) d x$
Now set $2 a - x = u$ and what happens if $f (2 a - x) = f (x)$

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