Shirley Peck

2023-03-23

Integration of 1/sinx-sin2x dx

### Answer & Explanation

nickalasaurus6ea0

Step 1
$\int \frac{1}{\mathrm{sin}x-\mathrm{sin}2x}\phantom{\rule{0ex}{0ex}}=-\int \frac{dx}{\mathrm{sin}2x-\mathrm{sin}x}\phantom{\rule{0ex}{0ex}}=\int \frac{\mathrm{sin}xdx}{\left(2\mathrm{cos}x-1\right)\left({\mathrm{cos}}^{3}x-1\right)}$
Step 2
Let u=cos x, du=-sin x dx
$=\int -\frac{du}{\left(2u-1\right)\left({u}^{2}-1\right)}=\int \frac{-du}{\left(u-1\right)\left(u+1\right)\left(2u-1\right)}\phantom{\rule{0ex}{0ex}}=-\left[\int -\frac{u}{3\left(2u-1\right)}+\frac{1}{6\left(u+1\right)}+\frac{1}{2\left(u-1\right)du}\right]\phantom{\rule{0ex}{0ex}}=-\left(\frac{2}{3}\mathrm{ln}\left(2u-1\right)+\frac{\mathrm{ln}\left(u+1\right)}{6}+\frac{\mathrm{ln}\left(u-1\right)}{2}\right)+c\phantom{\rule{0ex}{0ex}}=\frac{2}{3}\mathrm{ln}\left(2\mathrm{cos}x-1\right)-\frac{\mathrm{ln}\left(\mathrm{cos}x+1\right)}{6}-\frac{\mathrm{ln}\left(\mathrm{cos}x-1\right)}{2}+c$

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