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Question Number 85766 by jagoll last updated on 24/Mar/20

$$\frac{\mathrm{dy}}{\mathrm{dx}}=1-\sin (\mathrm{x}+2\mathrm{y})$$

Commented by jagoll last updated on 24/Mar/20

$$\mathrm{u}=\mathrm{x}+2\mathrm{y}$$$$\frac{\mathrm{du}}{\mathrm{dx}}=1+2\frac{\mathrm{dy}}{\mathrm{dx}}$$$$\frac{2\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dx}}-1\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}[\frac{\mathrm{du}}{\mathrm{dx}}-1]$$$$\Rightarrow \frac{1}{2}\frac{\mathrm{du}}{\mathrm{dx}}-\frac{1}{2}=1-\sin \mathrm{u}$$$$\frac{\mathrm{du}}{\mathrm{dx}}=3-2\sin \mathrm{u}\Rightarrow \frac{\mathrm{du}}{3-2\sin \mathrm{u}}=\mathrm{dx}$$$$\int \frac{\mathrm{du}}{3-2\sin \mathrm{u}}=\mathrm{x}+\mathrm{c}$$$$$$

Commented by john santu last updated on 25/Mar/20

$$\int \frac{du}{3-2\sin u}=\int \frac{2dt}{(1+t^2)(3-\frac{4}{1+t^2})}$$$$[u=\tan \frac{t}{2}]$$$$\int \frac{2dt}{3t^2-1}=\int \frac{dt}{\sqrt{3}t-1}-\int \frac{dt}{\sqrt{3}t+1}$$$$=\frac{1}{\sqrt{3}}\ln \mid \frac{\sqrt{3}t-1}{\sqrt{3}t+1}\mid =\frac{1}{\sqrt{3}}\ln \mid \frac{2\sqrt{3}{\tan }^{-1}\left(u\right)-1}{2\sqrt{3}{\tan }^{-1}\left(u\right)+1}\mid$$$$\therefore \frac{1}{\sqrt{3}}\ln \mid \frac{2\sqrt{3}{\tan }^{-1}(x+2y)-1}{2\sqrt{3}{\tan }^{-1}(x+2y)+1}\mid =x+c$$$$$$

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