Question

cos (x + y) dy = dx

Answer 1

We have,

\[\cos \left( x + y \right)dy = dx\]

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

Let `x + y = v`

\[ \Rightarrow 1 + \frac{dy}{dx} = \frac{dv}{dx}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{dv}{dx} - 1\]

Therefore, (1) becomes

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

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

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

Integrating both sides, we get

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

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

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

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

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

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

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

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

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

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

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

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

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