How do you evaluate int (sin^3x)^(1/2)(cosx)^(1/2) from [0,pi/2]?

1 Answer
Jan 22, 2018

int_0^(pi/2) (sinx)^(3/2)*(cosx)^(1/2)*dx=(pisqrt2)/8

Explanation:

I=int_0^(pi/2) (sinx)^(3/2)*(cosx)^(1/2)*dx

=int_0^(pi/2) (sinx/cosx)^(3/2)*(cosx)^2*dx

=int_0^(pi/2) (tanx)^(3/2)*(cosx)^2*dx

=int_0^(pi/2) (tanx)^(3/2)*(secx)^2/(secx)^4*dx

=2int_0^(pi/2) (tanx)^(2)/[(tanx)^2+1]^2*(secx)^2/(2sqrt(tanx))*dx

After using y=sqrt(tanx) and dy=(secx)^2/(2sqrt(tanx))*dx transforms, I became

I=int_0^oo (2y^4*dy)/(y^4+1)^2

=1/2int_0^oo y*(4y^3*dy)/(y^4+1)^2

=1/2*[y*-1/(y^4+1)]_0^oo-1/2int_0^oo -1/(y^4+1)*dy

=-1/2*[y/(y^4+1)]_0^oo+1/2int_0^oo dy/(y^4+1)

=1/2int_0^oo dy/(y^4+1)

After setting J=int_0^oo dy/(y^4+1), I must be equal to J/2

After using y=1/z and dy=-dz/z^2 transforms, J became

J=int_oo^0 (-dz/z^2)/((1/z)^4+1)

=int_oo^0 (-z^2*dz)/(z^4+1)

=int_0^oo (z^2*dz)/(z^4+1)

=int_0^oo (y^2*dy)/(y^4+1)

After collecting 2 integrals,

2J=int_0^oo ((y^2+1)*dy)/(y^4+1)

=1/2int_0^oo ((2y^2+2)*dy)/(y^4+1)

=1/2int_0^oo ((2y^2+2)*dy)/[(y^2+sqrt2*y+1)*(y^2-sqrt2*y+1)]

=1/2int_0^oo (dy)/(y^2+sqrt2*y+1)+1/2int_0^oo (dy)/(y^2-sqrt2*y+1)

=1/2int_0^oo (2dy)/(2y^2+2sqrt2*y+2)+1/2int_0^oo (2dy)/(2y^2-2sqrt2*y+2)

=sqrt2/2int_0^oo (sqrt2dy)/((sqrt2*y+1)^2+1)+sqrt2/2int_0^oo (sqrt2dy)/((sqrt2*y-1)^2+1)

=sqrt2/2*[arctan(sqrt2*y+1)]_0^oo+sqrt2/2*[arctan(sqrt2*y-1)]_0^oo

=(pisqrt2)/2

Thus,

I=J/2=(pisqrt2)/8