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Question Number 60637 by rajesh4661kumar@gamil.com last updated on 23/May/19

Commented by maxmathsup by imad last updated on 23/May/19

$$\int \frac{sinx}{1-sinx}dx=-\int \frac{1-sinx-1}{1-sinx}dx=-x+\int \frac{dx}{1-sinx}$$$$\int \frac{dx}{1-sinx}{=}_{tan\left(\frac{x}{2}\right)=t}\int \frac{1}{1-\frac{2t}{1+t^2}}\frac{2dt}{1+t^2}=2\int \frac{dt}{1+t^2-2t}=2\int \frac{dt}{{(t-1)}^2}$$$$=\frac{-2}{t-1}+c=\frac{2}{1-tan\left(\frac{x}{2}\right)}+c\Rightarrow$$$$\int \frac{sinx}{1-sinx}dx=-x+\frac{2}{1-tan\left(\frac{x}{2}\right)}+c.$$

Answered by tanmay last updated on 23/May/19

$$\int \frac{sinx(1+sinx)}{{cos}^{2}x}dx$$$$\int \frac{sinx}{{cos}^{2}x}dx+\int \frac{{sin}^{2}x}{{cos}^{2}x}dx$$$$\int tanxsecxdx+\int ({sec}^{2}x-1)dx$$$$\int tanxsecx+\int {sec}^{2}xdx-\int dx$$$$secx+tanx-x+c$$$$$$

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