# Given f(x)=arcsinx, how show integral (0-1)xf(x)dx <= pie/4?

sciencesolve | Certified Educator

You need to test for validity the statement `int_0^1 x*f(x) dx <= pi/4` , hence, you need to consider the following inequality such that:

`If x in [0,1] => x*f(x) < 1*f(x)`

Integrating both sides yields:

`int_0^1 x*f(x) dx <int_0^1 f(x) dx`

The problem provides the equation of function `f(x) = sin^(-1) x` , hence, you need to replace `f(x)` by its equation, such that:

`int_0^1f(x) dx =int_0^1 sin^(-1) x dx`

You need to use integration by parts such that:

`u = sin^(-1) x => du = 1/(sqrt(1 - x^2))`

`dv = 1 => v = x`

`int_0^1 sin^(-1) x dx = x*sin^(-1) x|_0^1 - int_0^1 x/(sqrt(1 - x^2)) dx`

You should notice that `-x/(sqrt(1 - x^2)) `  represents the derivative of `sqrt(1 - x^2)` such that:

`(sqrt(1 - x^2))' = -(2x)/(2sqrt(1 - x^2))`

Replacing `(sqrt(1 - x^2))'` for `-(2x)/(2sqrt(1 - x^2))` yields:

`int_0^1 sin^(-1) x dx = x*sin^(-1) x|_0^1 + int_0^1 (sqrt(1 - x^2))' dx`

Using the relation `int f'(x) dx = f(x)` yields:

`int_0^1 sin^(-1) x dx = x*sin^(-1) x|_0^1 + sqrt(1 - x^2)|_0^1`

Using the fundamental theorem of calculus yields:

`int_0^1 sin^(-1) x dx = 1*sin^(-1) 1 - 0*sin^(-1) 0 + sqrt(1 - 1^2) - sqrt(1 - 0^2)`

`int_0^1 sin^(-1) x dx = pi/2 - 0 + 0 - 1`

`int_0^1 sin^(-1) x dx = pi/2 - 1`

Comparing `pi/2 - 1` to `pi/4` yields:

`pi/2 - 1 < pi/4 => pi/2 - pi/4 < 1 => pi/4 < 1 => pi < 4` valid

Since `int_0^1 sin^(-1) x dx = pi/2 - 1 < pi/4` , the validity of statement  `int_0^1 x*f(x) dx <= pi/4` is checked.

Hence, evaluating the validity of the statement` int_0^1 x*f(x) dx <= pi/4` yields that it holds.

oldnick | Student

`int_0^1 x arcsinx dx=[x^2 arc sinx +xsqrt(1-x^2)-1/2xsqrt(1-x^2)-1/2 arc sinx]_0^1-`

`-int_0^1 x arcsinx dx`

So:

` int_0^1 x arcsinx dx=1/2[x^2arc sinx +1/2xsqrt(1-x^2)-1/2arc sin x]_0^1=`

`=1/2[pi/2-pi/4-0]=pi/8< pi/4`

oldnick | Student

`int arc sinx dx=x arcsin x-int (x dx)/sqrt(1-x^2)=` `=x arcsin x+ sqrt(1-x^2)+c`

On the other side:

`int sqrt(1-x^2) dx`

setting:  `x=sin z`  ; `dx=cosz dz`

`int sqrt(1-x^2) dx= int cos^2 z dz=1/2int (cos2z+1)dz=`

`=1/4 sin 2z +z/2+C=` `1/2[x sqrt(1-x^2)+arcsin x+C]`

Now developing by parts integral:

`int_0^1 x arcsinx dx=[x(x arcsinx+sqrt(1-x^2))]_0^1-`

`int_0^1 x arcsinx dx- int_0^1 sqrt(1-x^2)dx=`

TO BE CONTINUE..................