Question

Evaluate: $\int \frac{x+\sin x}{1-\cos x}\ dx$

Answer 1

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here is the solution,

integral ex((1-sinx) / (1-cosx))

=integral ex ((1-2sinx/2 cosx/2) / (2sin2x/2)  since 1-cosx=2sin2x/2

=integral -ex((-1/2 cosec2x/2 + cotx/2))

note it is of the form integral ex(( F(x) + F'(x))) where F(x) = cotx/2

this is a standard result whose solution is exF(x) +c  (for proof apply uv rule for exF'(x))

= -excotx/2 ........ answer

hope you are clear with the solution...

Answer 2

Solution:

$\int \frac{x+sinx}{1-cos x}dx = \int \frac{x}{1-cos x} dx+\int \frac{sin x}{1-cos x} dx... eq1$

As we know that

$1- cos x = 2 sin^{2}(\frac{x}{2})\\\\sin x= 2 sin \frac{x}{2}cos \frac{x}{2}$

Put this value in the given function to convert in integrable form

$\int\frac{x}{2 sin^{2}\frac{x}{2}}dx\\\\=\frac{1}{2}\int x\:cosec ^{2}(\frac{x}{2})dx$

Now integrate by parts

$\int x\:cosec ^{2}(\frac{x}{2})dx=x\int cosec ^{2}(\frac{x}{2})dx -\int[ \frac{dx}{dx}\int cosec ^{2}(\frac{x}{2})dx]dx\\\\=-2x cot(\frac{x}{2})+2\int cot(\frac{x}{2})dx\\\\\int \frac{x}{1-\cos x}\ dx= -2x cot(\frac{x}{2})+4\: log\:sin(\frac{x}{2})+C...eq2$

Now integrate second term

$\int \frac{2\:sin \frac{x}{2} cos\:(\frac{x}{2})}{2 sin^{2}(\frac{x}{2})}dx \\\\=\int cot(\frac{x}{2} ) dx \\\\=2log(sin\frac{x}{2})+C...eq3$

Put eq 2 and 3 in eq1 to get final result

$\int \frac{x+sinx}{1-cos x}dx =-2x cot(\frac{x}{2})+4\: log\:sin(\frac{x}{2})+2log(sin\frac{x}{2}+C\\\\= -2x cot(\frac{x}{2})+6\: log\:sin(\frac{x}{2})+C$