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Question Number 85503 by john santu last updated on 22/Mar/20

$$\frac{dy}{dx}=\sec (x+y)$$

Answered by mr W last updated on 22/Mar/20

$$u=x+y$$$$\frac{du}{dx}=1+\frac{dy}{dx}$$$$\frac{du}{dx}-1=\frac{1}{\cos u}$$$$\frac{\cos udu}{1+\cos u}=dx$$$$\int \frac{\cos udu}{1+\cos u}=\int dx$$$$\int (1-\frac{1}{1+\cos u})du=\int dx$$$$u-\tan \frac{u}{2}=x+C$$$$\Rightarrow y-\tan \frac{x+y}{2}=C$$

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