Question

dy/dx=cos(x+y)+sin(x+y)​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\blue{\sf{\dfrac{dy}{dx}=cos(x+y)+sin(x+y)}}$

To solve this, first

$\green{\tt{Put\,\,\,x+y=v}}$

Now, differentiate both sides w.r.t x

$\green{\tt{\mapsto\,1+\dfrac{dy}{dx}=\dfrac{dv}{dx}}}$

$\green{\tt{\mapsto\,\dfrac{dy}{dx}=\dfrac{dv}{dx}-1}}$

Substitute these into our original equation,

$\sf{\dfrac{dv}{dx}-1=cos(v)+sin(v)}$

$\sf{\implies\,\dfrac{dv}{dx}=cos(v)+sin(v)+1}$

$\sf{\implies\,\dfrac{dv}{cos(v)+sin(v)+1}=dx}$

Integrating both sides,

$\displaystyle\sf{\implies\,\int\dfrac{dv}{cos(v)+sin(v)+1}=\int\,dx}$

We know,

$\boxed{\bullet\,\,\,\,\purple{\bf{sin(\theta)=\dfrac{2\,\,tan\left(\dfrac{\theta}{2}\right)}{1+tan^2\left(\dfrac{\theta}{2}\right)}\,\,\,\,\,\&\,\,\,\,\,cos(\theta)=\dfrac{1-tan^2\left(\dfrac{\theta}{2}\right)}{1+tan^2\left(\dfrac{\theta}{2}\right)}}}}$

So,

$\displaystyle\sf{\implies\,\int\dfrac{dv}{\dfrac{1-tan^2\left(\dfrac{v}{2}\right)}{1+tan^2\left(\dfrac{v}{2}\right)}+\dfrac{2\,\,tan\left(\dfrac{v}{2}\right)}{1+tan^2\left(\dfrac{v}{2}\right)}+1}=\int\,dx}$

$\displaystyle\sf{\implies\,\int\dfrac{dv}{\dfrac{1-tan^2\left(\dfrac{v}{2}\right)+2\,\,tan\left(\dfrac{v}{2}\right)+1+tan^2\left(\dfrac{v}{2}\right)}{1+tan^2\left(\dfrac{v}{2}\right)}}=\int\,dx}$

$\displaystyle\sf{\implies\,\int\dfrac{1+tan^2\left(\dfrac{v}{2}\right)}{1-tan^2\left(\dfrac{v}{2}\right)+2\,\,tan\left(\dfrac{v}{2}\right)+1+tan^2\left(\dfrac{v}{2}\right)}\,dv=\int\,dx}$

$\displaystyle\sf{\implies\,\int\dfrac{sec^2\left(\dfrac{v}{2}\right)}{1+2\,\,tan\left(\dfrac{v}{2}\right)+1}\,dv=\int\,dx}$

$\displaystyle\sf{\implies\,\int\dfrac{sec^2\left(\dfrac{v}{2}\right)}{2+2\,\,tan\left(\dfrac{v}{2}\right)}\,dv=\int\,dx}$

$\displaystyle\sf{\implies\,\int\dfrac{sec^2\left(\dfrac{v}{2}\right)}{2\left\{1+tan\left(\dfrac{v}{2}\right)\right\}}\,dv=\int\,dx}$

$\displaystyle\sf{\implies\,\int\dfrac{\dfrac{1}{2}\cdot sec^2\left(\dfrac{v}{2}\right)}{1+tan\left(\dfrac{v}{2}\right)}\,dv=\int\,dx}$

In the integral in LHS,

$\green{\tt{Put\,\,\,1+tan\left(\dfrac{v}{2}\right)=u}}$

$\green{\tt{\mapsto\,\dfrac{1}{2}\cdot sec^2\left(\dfrac{v}{2}\right)\,dv=du}}$

So,

$\displaystyle\sf{\implies\,\int\dfrac{du}{u}=\int\,dx}$

$\displaystyle\sf{\implies\,\ln|u|=x+c}$

Substitute the value of u,

$\displaystyle\sf{\implies\,\ln\left|1+tan\left(\dfrac{v}{2}\right)\right|=x+c}$

Substitute the value of v,

$\displaystyle\sf{\implies\,\ln\left|1+tan\left(\dfrac{x+y}{2}\right)\right|=x+c}$