Question

Evaluate `int_0^1 x^5 sin ^-1 x dx`and find the value of β `(9/2,1/2)` 

Answer 1

`I= int_0^1 x^5 sin^-1 x dx` 

Put `sin^-1 x= t`          ∴ `x= sin t  dx= cost dt` 

When x = 0, t = 0      When` x=1, t=pi/2` 

`I= int_0^pi/2 sin^5t.t. cos t  dt = int_0^pi/2 t (sin^5t. cos t )dt` 

Integrating by parts, 

`1I=[t. sin^6 x/6]^(pi/2)-int_0^(pi/2) sin^6 x/6. 1.dt `

`I=(pi/2. 1/6-0)-1/6. (5.3.1)/(6.4.2).pi/2` 

 `I= pi/12-(5pi)/192` 

∴` I= (11pi)/192`  

`β (9/2,1/2)=(|~9/2|~1/2)/(|~5)= (7/2. 5/2. 3/2. 1/2. |~1/2 |~1/2)/(5.4.3.2.1)`

`β (9/2,1/2)=(105 pi)/384`