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Question Number 64410 by mmkkmm000m last updated on 17/Jul/19

$$\int 1/(1+ysin\theta )d\theta$$

Commented by mr W last updated on 17/Jul/19

$$seeQ64238sir!$$

Commented by Tony Lin last updated on 17/Jul/19

$$\int \frac{d\theta }{1+ysin\theta }$$$$=\int \frac{d\theta }{1+\frac{2ytan\frac{\theta }{2}}{{tan}^2\frac{\theta }{2}+1}}$$$$letu=tan\frac{\theta }{2}$$$$\Rightarrow \frac{du}{d\theta }=\frac{{sec}^2\frac{\theta }{2}}{2}$$$$\Rightarrow 2\int \frac{du}{u^2+2yu+1}$$$$=2\int \frac{du}{{(u+y)}^2-y^2+1}$$$$=2\int \frac{du}{(u+y-\sqrt{y^2-1})(u+y+\sqrt{y^2-1})}$$$$=\frac{1}{\sqrt{y^2-1}}(\int \frac{du}{u+y-\sqrt{y^2-1}}-\int \frac{du}{u+y+\sqrt{y^2-1}})$$$$=\frac{1}{\sqrt{y^2-1}}(ln\mid u+y-\sqrt{y^2-1}\mid -ln\mid u+y+\sqrt{y^2-1}\mid +c$$$$=\frac{ln\mid tan\frac{\theta }{2}+y-\sqrt{y^2-1}\mid -ln\mid tan\frac{\theta }{2}+y+\sqrt{y^2-1}\mid }{\sqrt{y^2-1}}+c$$

Answered by Tony Lin last updated on 17/Jul/19

$$theformofyourpostsaretoobig\cdot \cdot \cdot \cdot \cdot \cdot$$

Commented by mr W last updated on 17/Jul/19

$$isuggestyounottoanswerthosequestions$$$$beforesomeonechanges.$$

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