Prove the integral inequality π 4 − 1 2 ≤ ∫ 0 1 arcsin ⁡ x 1 + x 8 ≤ π 2 − 1 π 4 − 1 2 ≤ ∫ 0 1 arcsin ⁡ x 1 + x 8 ≤ π 2 − 1I can say that I have close to zero experience with integral inequalities. I've searched for couple of theorems most of them involving monotonous function on [0,1]. Arcsin is increasing in this case and the other one is decreasing. You could say we have a product of increasing and decreasing function on this interval however I didn't find any theorems that cover this case. Also this integral doesn't seem to have elementary function.Another thing I noticed is that most of the theorems cover only 1 part of this inequality; either ≤ o r ≥ so my guess is we need some type of combination or perhaps find a new integral that is equal to π 4 − 1 2 and another that is equal to π 2 − 1.What's the correct approach when we need to prove that the value of certain integral is in a provided interval?

Question

Prove the integral inequality      π    4    −      1    2    ≤      ∫    0    1                      arcsin        ⁡        x                    1        +                  x          8                      ≤      π    2    −  1      π    4    −      1    2    ≤      ∫    0    1                      arcsin        ⁡        x                    1        +                  x          8                      ≤      π    2    −  1I can say that I have close to zero experience with integral inequalities. I&#039;ve searched for couple of theorems most of them involving monotonous function on [0,1]. Arcsin is increasing in this case and the other one is decreasing. You could say we have a product of increasing and decreasing function on this interval however I didn&#039;t find any theorems that cover this case. Also this integral doesn&#039;t seem to have elementary function.Another thing I noticed is that most of the theorems cover only 1 part of this inequality; either   ≤  o  r  ≥ so my guess is we need some type of combination or perhaps find a new integral that is equal to       π    4    −      1    2   and another that is equal to       π    2    −  1.What&#039;s the correct approach when we need to prove that the value of certain integral is in a provided interval?

Ricky Lamb · Accepted Answer

Step 1For the bound on the right,       ∫    0    1                      arcsin        ⁡        x                    1        +                  x          8                        d  x  ≤      ∫    0    1    arcsin  ⁡  x    d  x  =      π    2    −  1..Step 2For the bound on the left,       ∫    0    1                      arcsin        ⁡        x                    1        +                  x          8                        d  x  ≥      1    2          ∫          0        1              arcsin      ⁡      x          d  x  =      1    2          (          π      2        −    1    )    =      π    4    −      1    2    .

Prove the integral inequality pi/4-1/2 leq int_0^1 arcsin/(1+x^8) leq pi/2-1.

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