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Question Number 149081 by Integrals last updated on 02/Aug/21

Answered by Ar Brandon last updated on 02/Aug/21

$$\varphi =\int \frac{d\beta }{2-3\sin \beta },t=\tan \frac{\beta }{2}$$$$=\int \frac{2}{2-3\left(\frac{2t}{1+t^2}\right)}\cdot \frac{dt}{1+t^2}=\int \frac{dt}{t^2-3t+1}$$$$=\int \frac{dt}{{(t-\frac{3}{2})}^2-\frac{5}{4}}=-\frac{2}{\sqrt{5}}\mathrm{arctanh}\left(\frac{2t-3}{\sqrt{5}}\right)+C$$$$=\frac{1}{\sqrt{5}}\ln \mid \frac{2t-3-\sqrt{5}}{2t-3+\sqrt{5}}\mid +C=\frac{\sqrt{5}}{5}\ln \mid \frac{2\tan \left(\frac{\beta }{2}\right)-3-\sqrt{5}}{2\tan \left(\frac{\beta }{2}\right)-3+\sqrt{5}}\mid +C$$

Commented by Ar Brandon last updated on 02/Aug/21

$$\sin \beta =\frac{\sin \beta }{1}=\frac{2\sin \frac{\beta }{2}\cos \frac{\beta }{2}}{{\cos }^2\frac{\beta }{2}+{\sin }^2\frac{\beta }{2}}=\frac{2\tan \frac{\beta }{2}}{1+{\tan }^2\frac{\beta }{2}}$$$$t=\tan \frac{\beta }{2}\Rightarrow dt=\frac{1}{2}{\sec }^2\frac{\beta }{2}d\beta =\frac{1}{2}(1+{\tan }^2\frac{\beta }{2})d\beta$$$$\mathrm{arctanh}\left(x\right)=\frac{1}{2}\ln \mid \frac{1+x}{1-x}\mid$$

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