Question

sinx /(1-cos x)(2-cosx) integration

Answer 1

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Answer 2

Answer:

$\int \frac{sin \: x}{(1 - cos \: x)(2 - cos \: x)} dx = log\bigg(\bigg | \frac{cos \: x - 1}{cos \: x - 2} \bigg| \bigg) + c \\ \\ \:$

Step-by-step explanation:

$\int \frac{sin \: x}{(1 - cos \: x)(2 - cos \: x)} dx$

By inspecting one can say we can substitute cos x as t

Let

$cos \: x = t \\ \\ - sin \: x \: dx = dt \\ \\ do \: substitution \\ \\ \int \frac{ - dt}{(1 - t)(2 - t)} dx$

Now,do partial fraction

$\frac{ - 1}{(1 - t)(2 - t)} = \frac{A}{1 - t} + \frac{B}{2 - t} \\ \\ - 1 = A(2 - t) + B(1 - t) \\ \\ - 1 = 2A + B - At - Bt \\ \\ - 1 = 2A + B+ t( - A - B) \\ \\ compare \: the \: coefficient \\ \\ 2A+ B= - 1 \\ - A - B = 0 \\ \\ A = - 1 \\ \\ B = 1$

So

$\int\frac{ - 1}{1 - t} dt+ \int\frac{1}{2 - t}dt \\ \\or \\ \\ \int\frac{ 1}{ t - 1} dt - \int\frac{1}{t - 2}dt = log( |t - 1| ) - log( |t - 2| ) + c \\ \\ \\ \int\frac{ 1}{ t - 1} dt - \int\frac{1}{t - 2}dt= log\bigg( \bigg| \frac{t - 1}{t - 2} \bigg| \bigg) + c$

undo substitution

$\int \frac{sin \: x}{(1 - cos \: x)(2 - cos \: x)} dx = log\bigg(\bigg | \frac{cos \: x - 1}{cos \: x - 2} \bigg| \bigg) + c \\ \\ \:$

Hope it helps you.