Integrate 1/(sin x.cos^3 x)

please help me solve this trouble!!!!!!!

∫ 1 / (sinx cos^3 x) dx
= ∫ (cos^2 x + sin^2 x) dx / (sinx cos^3 x)
= ∫ dx / (sinx cosx) + ∫ sinx dx / cos^3 x
= 2 ∫ cosec (2x) dx + ∫ sinx dx / cos^3 x
= 2 * (1/2) ln l cosec2x - cot2x l + ∫ sinx dx / cos^3 x
= ln l cosec2x - cot2x l + ∫ sinx dx / cos^3 x
= ln l (1 - cos2x) / sin2x l + ∫ sinx dx / cos^3 x
= ln l tanx l + ∫ sinx dx / cos^3 x

For the second integral, let cosx = t
=> - sinx dx = dt
=> ∫ sinx dx / cos^3 x
= - ∫ dt / t^3
= t^2 / 2 + c
= cos^2 x / 2 + c
= (1/2) sec^2 x + c

=> Reqd. integral
= ln l tanx l + (1/2) sec^2 x + c.

Integration answers may look different, but all correct answers differ by a constant. If you give me your book answer, I shall prove that both the answers are correct.

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∫1/(sin x * cos^3 x) dx

∫1/(sin x * cos x * cos² x) dx

∫1/(sin(2x)/2 * (1 + cos(2x))/2) dx

4* ∫dx/(sin(2x))(1 + cos(2x))

4* ∫[(1 - cos(2x)]/[sin^3(2x)]

4*∫csc^3(2x) dx - 4*∫cos(2x)/sin^3(2x) dx

The first integral can be done by parts and second integral requires a substitution.