How to solve this ∫ sinx/[(1-cosx)(2-cosx)].dx pls help me...?

take cos x = t
Differentiate both sides
- sin x dx = dt
The Integral boils down to
-∫ dt / (1-t)(2-t)
Do partial fractions
∫ ( 1/ t-1) - (1/ 2-t) dt
= log (t-1) + log(t-2) + c
Substitute t = cos x
Answer: log(cos x - 1) + log(cos x -2 ) + c

L=S(a million-sinx)/cosxdx =S(a million/cosx-sinx/cosx)dx =S(cosx/a million-sin ^2 x)dx -S (sinx/cosx)dx= J - ok * J: U=sinx => du=cosxdx =>j=S du/(a million-u2) =S du/(a million-u)(a million+u)= a million/2S du *(a million/a million-u + a million/a million+u) =a million/2ln I a million+u/a million-u I + const = a million/2 ln I a million+sinx/a million-sinx I + const *ok ok=S -(cosx')/cosx dx= -ln IcosxI +const =>L

∫ sinx/[(1-cosx)(2-cosx)].dx
Let 1−cosx= z→sinxdx=dz →dx=dz/sinx
and 2−cosx=1+1−cosx=1+z
Hence
∫ sinx/[(1-cosx)(2-cosx)].dx = ∫ sinx/[z(1+z)].dz/sinx
= ∫ 1/[z(1+z)].dz
Break up 1/[z(1+z)] into partial fractions
1/[z(1+z)]= A/z+B/(1+z) =[A(1+z)+Bz]/[z(1+z)] =[A+(A+B)z]/[z(1+z)]
Hence A= 1 and A+B=0→B=−A=−1
Hence 1/[z(1+z)] =1/z−1/(1+z)
and = ∫ 1/[z(1+z)].dz== ∫ [1/z−1/(1+z)].dz
=lnz −ln(1+z) +C=ln(1−cosx) −ln(1+1−cosx)+C
∫ sinx/[(1-cosx)(2-cosx)].dx=ln(1−cosx) −ln(2−cosx)+C

1 )u substitution ...u = cos x to obtain integral ( -1/ (1-u)(2-u))

then use partial fractions , then integrate the result which should be easy

then back substitute for u with cos x

Let u = 1 - cosx.
dx = du/(sinx)

∫ sinx/[(1 - cosx)(2 - cosx)] dx
= ∫ sinx/[u(1 + u)] du/(sinx)
= ∫ 1/[u(u + 1)] du

1 = Au + B(u + 1)
1 = (A + B)u + B
B = 1 and A = -1.

= ∫ 1/u - 1/(u + 1) du
= ln|u| - ln|u + 1| + C
= ln|1 - cosx| - ln|2 - cosx| + C where C is a constant.