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Question Number 163730 by HongKing last updated on 09/Jan/22

$$\mathrm{Find}:\mathbf{\Omega }=\int \frac{\sin \left(\mathrm{x}\right)+\sqrt{3}\cos \left(\mathrm{x}\right)}{\sin \left(3\mathrm{x}\right)}\mathrm{dx}$$

Answered by cortano1 last updated on 10/Jan/22

$$=2\int \frac{\sin (x+\frac{\pi }{3})}{\sin \left(3x\right)}dx$$$$=2\int \frac{d\left(\tan \theta \right)}{3-\tan {}^2\theta };\theta =x+\frac{\pi }{6}$$$$=\frac{2\sqrt{3}}{3}\tanh {}^{-1}\left(\frac{\tan \theta }{\sqrt{3}}\right)+c$$$$=\frac{2\sqrt{3}}{3}\tanh {}^{-1}\left(\frac{\tan (x+\frac{\pi }{6})}{\sqrt{3}}\right)+c$$

Answered by MJS_new last updated on 10/Jan/22

$$\int \frac{\sin x+\sqrt{3}\cos x}{\sin 3x}dx=$$$$=-\int \frac{(\sqrt{3}+\tan x)(1+{\tan }^{2}x)}{(3-\tan x)\tan x}dx=$$$$=\int \frac{1+{\tan }^{2}x}{(\sqrt{3}-\tan x)\tan x}dx=$$$$[t=\tan x\to dx=\frac{dt}{1+{\tan }^{2}x}]$$$$=\int \frac{dt}{t(\sqrt{3}-t)}=\frac{\sqrt{3}}{3}\ln \mid \frac{t}{t-\sqrt{3}}\mid =$$$$=\frac{\sqrt{3}}{3}\ln \mid \frac{\sin x}{\sin x-\sqrt{3}\cos x}\mid +C$$

Commented by peter frank last updated on 11/Jan/22

$$\mathrm{great}$$

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