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$$\int \frac{2 \sin x+ \sin 2x}{1 - \cos x }\mathrm{d}x . $$

the following to is to be solved by substitution method. I tried to substitute $1-\cos x$, but couldn't reach the solution. Any help is appreciated.

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  • $\begingroup$ Did you try$u = \cos x$? Note $\sin 2x = 2\sin x \cos x$. $\endgroup$ – user99914 Jan 7 '16 at 11:38
  • $\begingroup$ @JohnMa No i didn't. But after you said I tried but couldn't do it. If u can show how to do it,that would be great! $\endgroup$ – user302630 Jan 7 '16 at 11:43
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$$\begin{align}\int \frac{2 \sin x+ \sin 2x}{1 - \cos x }\ dx &= \int \frac{2 \sin x+ 2\sin x\cos x}{1 - \cos x }\ dx \\ &= 2\int\frac{1+\cos x}{1-\cos x}\sin x\ dx \\ &= 2\int\frac{-(1-\cos x)+2}{1-\cos x}\frac{d(1-\cos x)}{dx}\ dx \\ &= 2\int\frac{-(1-\cos x)+2}{1-\cos x}\ {d(1-\cos x)}\end{align}$$

Does that help?

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  • $\begingroup$ The answer is 4 log(1-cos x) + 2 cos x + c. Can you show how that is coming???? $\endgroup$ – user302630 Jan 7 '16 at 11:56
  • $\begingroup$ Replacing $1-\cos x$ with $u$ everywhere in that last expression gets you $2\int \frac{-u+2}{u}\ du$. Now see if you can evaluate it. $\endgroup$ – user137731 Jan 7 '16 at 11:58
  • $\begingroup$ Yeah yeah!!. My bad. Got it . Thanks!! $\endgroup$ – user302630 Jan 7 '16 at 12:04
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Hint

$$I=\int \frac{2 \sin( x)+ \sin (2x)}{1 - \cos (x) }\ dx = \int \frac{2 \sin (x)+ 2\sin ( x)\cos ( x)}{1 - \cos ( x) }\ dx = 2\int\frac{1+\cos( x)}{1-\cos( x)}\sin ( x)\ dx $$

Now, change variable $y=\cos(x)$, $dy=-\sin(x)\,dx$ which makes $$I=-2\int\frac{1+y}{1-y}\,dy$$ which seems easy to solve.

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  • $\begingroup$ Can you please show the last few step and get the answer. I am having problem to do solve the last expression of yours $\endgroup$ – user302630 Jan 7 '16 at 11:53
  • $\begingroup$ @user302630 Set $u = 1 - y \implies du = - dy$ and $1 + y = 2 - u$, so your integral becomes $$2 \int \frac{2 - u}{u} du = 2 \int \frac{2}{u} - 1 du$$ $\endgroup$ – mattos Jan 7 '16 at 11:58
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$$u=1-\cos t$$ $$\int \frac{2 \sin x+ 2\sin x \cos x}{1 - \cos x }\ dx=2\int {2-u \over u} du$$

$$=\int (\frac4u-2) du=-2u+4\log u +C$$ $$=-2(1-\cos t)+4\log (1-\cos t) +C$$ $$= 2 \cos t +4\log (1-\cos t) +C'$$

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