Solved: "calculate this integral: ∫0π2arccos⁡(sin⁡x)dx the answer is −d{π}28..."

Aubrey Hendricks

Answered question

2022-01-23

calculate this integral:

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

the answer is

- d \frac{{π}^{2}}{8}

It doesnt

vasselefa

Beginner2022-01-24Added 9 answers

Yes, your result is correct. For

x \in [- 1, 1]

\arccos (x) = \frac{π}{2} - \arcsin (x)

Hence

\int_{0}^{\frac{π}{2}} \arccos (\sin (x)) d x = \int_{0}^{\frac{π}{2}} (\frac{π}{2} - x) d x = \int_{0}^{\frac{π}{2}} t d t = {[\frac{t^{2}}{2}]}_{0}^{\frac{π}{2}} = \frac{π^{2}}{8}

Moreover

t \to \arccos (t)

is positive in [−1,1) so the given integral has to be POSITIVE!

Jacob Trujillo

Beginner2022-01-25Added 13 answers

Essentially the same idea of the other answer, but using the covs

x = t + \frac{π}{2}

t=-s and the evenness of

\cos

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

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

\int_{- \frac{π}{2}}^{0} \arccos (\cos (t)) d t =

\int_{0}^{\frac{π}{2}} (\arccos (\cos (- s)) d s =

\int_{0}^{\frac{π}{2}} (\arccos (\cos (s)) d s =

\int_{0}^{\frac{π}{2}} s, d s = \frac{π^{2}}{8}

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