I=int_0^(pi/2) (sinx)^(3/2)*(cosx)^(1/2)*dxI=∫π20(sinx)32⋅(cosx)12⋅dx
=int_0^(pi/2) (sinx/cosx)^(3/2)*(cosx)^2*dx∫π20(sinxcosx)32⋅(cosx)2⋅dx
=int_0^(pi/2) (tanx)^(3/2)*(cosx)^2*dx∫π20(tanx)32⋅(cosx)2⋅dx
=int_0^(pi/2) (tanx)^(3/2)*(secx)^2/(secx)^4*dx∫π20(tanx)32⋅(secx)2(secx)4⋅dx
=2int_0^(pi/2) (tanx)^(2)/[(tanx)^2+1]^2*(secx)^2/(2sqrt(tanx))*dx2∫π20(tanx)2[(tanx)2+1]2⋅(secx)22√tanx⋅dx
After using y=sqrt(tanx)y=√tanx and dy=(secx)^2/(2sqrt(tanx))*dxdy=(secx)22√tanx⋅dx transforms, II became
I=int_0^oo (2y^4*dy)/(y^4+1)^2I=∫∞02y4⋅dy(y4+1)2
=1/2int_0^oo y*(4y^3*dy)/(y^4+1)^212∫∞0y⋅4y3⋅dy(y4+1)2
=1/2*[y*-1/(y^4+1)]_0^oo-1/2int_0^oo -1/(y^4+1)*dy12⋅[y⋅−1y4+1]∞0−12∫∞0−1y4+1⋅dy
=-1/2*[y/(y^4+1)]_0^oo+1/2int_0^oo dy/(y^4+1)−12⋅[yy4+1]∞0+12∫∞0dyy4+1
=1/2int_0^oo dy/(y^4+1)12∫∞0dyy4+1
After setting J=int_0^oo dy/(y^4+1)J=∫∞0dyy4+1, II must be equal to J/2J2
After using y=1/zy=1z and dy=-dz/z^2dy=−dzz2 transforms, JJ became
J=int_oo^0 (-dz/z^2)/((1/z)^4+1)J=∫0∞−dzz2(1z)4+1
=int_oo^0 (-z^2*dz)/(z^4+1)∫0∞−z2⋅dzz4+1
=int_0^oo (z^2*dz)/(z^4+1)∫∞0z2⋅dzz4+1
=int_0^oo (y^2*dy)/(y^4+1)∫∞0y2⋅dyy4+1
After collecting 2 integrals,
2J=int_0^oo ((y^2+1)*dy)/(y^4+1)2J=∫∞0(y2+1)⋅dyy4+1
=1/2int_0^oo ((2y^2+2)*dy)/(y^4+1)12∫∞0(2y2+2)⋅dyy4+1
=1/2int_0^oo ((2y^2+2)*dy)/[(y^2+sqrt2*y+1)*(y^2-sqrt2*y+1)]12∫∞0(2y2+2)⋅dy(y2+√2⋅y+1)⋅(y2−√2⋅y+1)
=1/2int_0^oo (dy)/(y^2+sqrt2*y+1)+1/2int_0^oo (dy)/(y^2-sqrt2*y+1)12∫∞0dyy2+√2⋅y+1+12∫∞0dyy2−√2⋅y+1
=1/2int_0^oo (2dy)/(2y^2+2sqrt2*y+2)+1/2int_0^oo (2dy)/(2y^2-2sqrt2*y+2)12∫∞02dy2y2+2√2⋅y+2+12∫∞02dy2y2−2√2⋅y+2
=sqrt2/2int_0^oo (sqrt2dy)/((sqrt2*y+1)^2+1)+sqrt2/2int_0^oo (sqrt2dy)/((sqrt2*y-1)^2+1)√22∫∞0√2dy(√2⋅y+1)2+1+√22∫∞0√2dy(√2⋅y−1)2+1
=sqrt2/2*[arctan(sqrt2*y+1)]_0^oo+sqrt2/2*[arctan(sqrt2*y-1)]_0^oo√22⋅[arctan(√2⋅y+1)]∞0+√22⋅[arctan(√2⋅y−1)]∞0
=(pisqrt2)/2π√22
Thus,
I=J/2=(pisqrt2)/8I=J2=π√28
