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Question Number 37432 by MJS last updated on 13/Jun/18

$$\int \frac{d\alpha }{\frac{1}{2}\tan \alpha +2\cot \frac{\alpha }{2}}=?$$$$\int \frac{d\beta }{2\tan \frac{\beta }{2}+\frac{1}{2}\cot \beta }=?$$

Answered by ajfour last updated on 13/Jun/18

$$\int \frac{2\sin \alpha \cos \alpha d\alpha }{\sin {}^2\alpha +4\cos {}^2\alpha }=\int \frac{\sin 2\alpha d\alpha }{1+\frac{3}{2}(1+\cos 2\alpha )}$$$$=\frac{1}{3}\int \frac{3\sin 2\alpha d\left(2\alpha \right)}{5+3\cos 2\alpha }$$$$=-\frac{1}{3}\ln (5+3\cos 2\alpha )+c.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jun/18

$$2)t=tan\frac{\beta }{2}dt={sec}^2\frac{\beta }{2}\times \frac{1}{2}\times d\beta$$$$\int \frac{2dt}{1+t^2}\times \frac{1}{2t+\frac{1}{2}\times \frac{1-t^2}{2t}}$$$$\int \frac{2dt}{1+t^2}\times \frac{4t}{8t^2+1-t^2}$$$$8\int \frac{tdt}{1+t^2}\times \frac{1}{1+7t^2}k=t^{2}dk=2tdt$$$$4\int \frac{dk}{1+k}\times \frac{1}{1+7k}$$$$\frac{4}{6}\int \frac{(7+7k)-(1+7k)}{(1+k)(1+7k)}dk$$$$\frac{2}{3}\int \frac{7dk}{1+7k}-\frac{2}{3}\int \frac{dk}{1+k}$$$$\frac{2}{3}ln\mid \frac{1+7k}{1+k}\mid$$$$\frac{2}{3}ln\mid \frac{1+7t^2}{1+t^2}\mid$$$$\frac{2}{3}ln\mid \frac{1+7{tan}^2\beta }{1+{tan}^2\beta }\mid$$$$$$$$$$

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