integration of cos^3x/(sin x)^1/2

Question

please help me solve this

Answer 1

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)-∫sin^3/2(x)cos(x)dx.

Let A=∫dx/√sin(x) and B=∫sin^3/2(x)cos(x)dx.

Let u=sin(x), du=cos(x)dx, B=∫u^3/2du=⅖u^5/2=⅖sin^5/2(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/(2^n-1n!)])du for integer n>0.

A=2u+∑[(1×...×(2n-1))u⁴ⁿ⁺¹/(2^n-1n!)]/(4n+1)

J=A-B+C, where C is integration constant.

J=2√sin(x)+∑[(1×...×(2n-1))sin²ⁿ(x)√sin(x)/(2^n-1n!)]/(4n+1)-⅖sin²(x)√sin(x)+C.