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Question Number 64735 by mmkkmm000m last updated on 20/Jul/19

$$\int tan\theta /1\stackrel{-}{+}sin\theta d\theta$$

Commented by mr W last updated on 21/Jul/19

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Commented by mathmax by abdo last updated on 21/Jul/19

$$letI=\int \frac{tan\theta }{1+sin\theta }d\theta changementtan\left(\frac{\theta }{2}\right)=tgive$$$$I=\int \frac{\frac{2t}{1-t^2}}{1+\frac{2t}{1+t^2}}\frac{2dt}{1+t^2}=\int \frac{4tdt}{(1-t^2)(1+t^2+2t)}=-\int \frac{4t}{(t^2-1){(t+1)}^2}dt$$$$=-4\int \frac{4t}{(t-1){(t+1)}^3}dtletdecomposeF\left(t\right)=\frac{t}{(t-1){(t+1)}^3}$$$$F\left(t\right)=\frac{a}{t-1}+\frac{b}{t+1}+\frac{c}{{(t+1)}^2}+\frac{d}{{(t+1)}^3}$$$$a={lim}_{x\to 1}(t-1)F\left(t\right)=\frac{1}{8}$$$$d={lim}_{t\to -1}{(t+1)}^{3}F\left(t\right)=\frac{1}{2}\Rightarrow F\left(t\right)=\frac{1}{8(t-1)}+\frac{b}{t+1}+\frac{c}{{(t+1)}^2}+\frac{1}{2{(t+1)}^3}$$$${lim}_{t\to +\mathrm{\infty }}tF\left(t\right)=0=a+b\Rightarrow b=-\frac{1}{8}\Rightarrow$$$$F\left(t\right)=\frac{1}{8(t-1)}-\frac{1}{8(t+1)}+\frac{c}{{(t+1)}^2}+\frac{1}{2{(t+1)}^3}$$$$F\left(0\right)=0=-\frac{1}{8}-\frac{1}{8}+c+\frac{1}{2}=-\frac{1}{4}+\frac{1}{2}+c=\frac{1}{4}+c\Rightarrow c=-\frac{1}{4}\Rightarrow$$$$F\left(t\right)=\frac{1}{8(t-1)}-\frac{1}{8(t+1)}-\frac{1}{4{(t+1)}^2}+\frac{1}{2{(t+1)}^3}\Rightarrow$$$$\int F\left(t\right)dt=\frac{1}{8}ln\mid t-1\mid -\frac{1}{8}ln\mid t+1\mid +\frac{1}{4(t+1)}-\frac{1}{4{(t+1)}^2}+C$$$$=\frac{1}{8}ln\mid \frac{t-1}{t+1}\mid +\frac{1}{4(t+1)}-\frac{1}{4{(t+1)}^2}+C\Rightarrow$$$$I=-\frac{1}{2}ln\mid \frac{t-1}{t+1}\mid -\frac{1}{(t+1)}+\frac{1}{{(t+1)}^2}+C$$$$=-\frac{1}{2}ln\mid \frac{tan\left(\frac{\theta }{2}\right)-1}{tan\left(\frac{\theta }{2}\right)+1}\mid -\frac{1}{1+tan\left(\frac{\theta }{2}\right)}+\frac{1}{{(1+tan\left(\frac{\theta }{2}\right))}^2}+C.$$$$$$

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