Question

\[\frac{dy}{dx} = \sec\left( x + y \right)\]

Answer 1

We have,

\[\frac{dy}{dx} = \sec\left( x + y \right)\]

\[\frac{dy}{dx} = \frac{1}{\cos\left( x + y \right)}\]

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

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

\[ \Rightarrow \int\frac{\cos v\left( 1 - \cos v \right)}{1 - \cos^2 v}dv = \int dx\]

\[ \Rightarrow \int\frac{\cos v\left( 1 - \cos v \right)}{\sin^2 v}dv = \int dx\]

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

\[ \Rightarrow \int\left( \cot v\ cosec + v - \cot^2 v \right)dv = \int dx\]

\[ \Rightarrow \int\left( \cot v\ cosec\ v - {cosec}^2 v + 1 \right)dv = \int dx\]

\[ \Rightarrow - cosec\ v + \cot v + v = x + C\]

\[ \Rightarrow - cosec \left( x + y \right) + \cot \left( x + y \right) + x + y = x + C\]

\[ \Rightarrow - cosec \left( x + y \right) + \cot \left( x + y \right) + y = C\]

\[ \Rightarrow \frac{- 1 + \cos \left( x + y \right)}{\sin \left( x + y \right)} + y = C\]

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

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