Mathematics

Evaluate : $$\int \frac{1}{sinx-sin2x}dx$$


SOLUTION
$$I=\int \dfrac{1}{sinx-sin2x}dx$$

$$I=\int \dfrac{1}{sinx-2sinxcosx}dx$$

$$I=\int \dfrac{1}{sinx(1-2cosx)}dx$$

$$I=\int \dfrac{sinx}{sin^{2}x(1-2cosx)}dx$$

$$I=\int \dfrac{sinx}{(1-cos^{2}x)(1-2cosx)}dx$$

Let $$cosx=t$$
$$-sinxdx=dt$$
$$sinxdx=-dt$$

$$I=\int \dfrac{-dt}{(1-t^{2})(1-2t)}$$

$$I=\int \dfrac{-dt}{(1-t)(1+t)(1-2t)}$$

$$I=\int \dfrac{1}{(t-1)(t+1)(1-2t)}dt$$

Now,
$$\dfrac{1}{(t-1)(t+1)(1-2t)}=\dfrac{A}{t+1} +\dfrac{B}{t-1} +\dfrac{C}{1-2t}$$

Solving this, we get,
$$A=-\dfrac{1}{6}, B=-\dfrac{1}{2}, C=-\dfrac{4}{3}$$
Therefore,
$$I=\int \dfrac{A}{t+1}dt +\int \dfrac{B}{t-1}dt + \int \dfrac{C}{1-2t}dt$$

$$I=Alog(t+1)+Blog(t-1)-\dfrac{C}{2}log(1-2t)+C$$

$$I=\dfrac{-1}{6}log(cosx+1)-\dfrac{1}{2}log(cosx-1)+\dfrac{4}{6}log(1-2cosx)+C$$
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Subjective Medium Published on 17th 09, 2020
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