Question

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

Answer 1

We have,

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

$$\frac{dy}{dx}=\frac{1}{\cos (x+y)}$$

Let x + y = v

$$\Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}$$

$$\Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1$$

$$\therefore \frac{dv}{dx}-1=\frac{1}{\cos v}$$

$$\Rightarrow \frac{dv}{dx}=\frac{\cos v+1}{\cos v}$$

$$\Rightarrow \frac{\cos v}{\cos v+1}dv=dx$$

Integrating both sides, we get

$$\int \frac{\cos v}{\cos v+1}dv=\int dx$$

$$\Rightarrow \int \frac{\cos v(1-\cos v)}{1-{\cos }^{2}v}dv=\int dx$$

$$\Rightarrow \int \frac{\cos v(1-\cos v)}{{\sin }^{2}v}dv=\int dx$$

$$\Rightarrow \int \frac{\cos v-{\cos }^{2}v}{{\sin }^{2}v}dv=\int dx$$

$$\Rightarrow \int (\cot vcosec+v-{\cot }^{2}v)dv=\int dx$$

$$\Rightarrow \int (\cot vcosecv-{cosec}^{2}v+1)dv=\int dx$$

$$\Rightarrow -cosecv+\cot v+v=x+C$$

$$\Rightarrow -cosec(x+y)+\cot (x+y)+x+y=x+C$$

$$\Rightarrow -cosec(x+y)+\cot (x+y)+y=C$$

$$\Rightarrow \frac{-1+\cos (x+y)}{\sin (x+y)}+y=C$$

$$\Rightarrow -\tan \left(\frac{x+y}{2}\right)+y=C$$

$$\Rightarrow y=\tan \left(\frac{x+y}{2}\right)+C$$