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

Question

Elijah Abram · Accepted Answer

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

#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#

Christopher Allers · Answer

undefined To evaluate $\int_{0}^{\frac{\pi}{2}} (\sin^3(x))^{1/2}(\cos(x))^{1/2} \, dx$, use the substitution $u = \sin(x)$ or $u = \cos(x)$, and then apply appropriate trigonometric identities. After substitution, the integral reduces to a form that can be evaluated more easily.