Let J=∫cos3(x)dx/√sin(x)=∫cos2(x).cos(x)dx/√sin(x),
J=∫(1-sin2(x))cos(x)dx/√sin(x)
J=∫dx/√sin(x)-∫sin3/2(x)cos(x)dx.
Let A=∫dx/√sin(x) and B=∫sin3/2(x)cos(x)dx.
Let u=sin(x), du=cos(x)dx, B=∫u3/2du=⅖u5/2=⅖sin5/2(x)=⅖sin2(x)√sin(x).
Let u2=sin(x), 2udu=cos(x)dx=√(1-u4)dx, dx=2udu/√(1-u4),
A=∫[2udu/√(1-u4)]/u=2∫du/√(1-u4).
A can be evaluated as an infinite series:
A=2∫(1-u4)-½du,
A=2∫(1+u4/2+(-1/2)(-3/2)u8/2!+(-1/2)(-3/2)(-5/2)u12/3!+...+(1.3.5...(2n-1))u4n/(2nn!)+...)du,
A=∫(2+∑[(1×...×(2n-1))u4n/(2n-1n!)])du for integer n>0.
A=2u+∑[(1×...×(2n-1))u4n+1/(2n-1n!)]/(4n+1)
J=A-B+C, where C is integration constant.
J=2√sin(x)+∑[(1×...×(2n-1))sin2n(x)√sin(x)/(2n-1n!)]/(4n+1)-⅖sin2(x)√sin(x)+C.