Question

Solve:

$\cos (x+y)dy=dx$

Answer 1

$cos(x+y)dy=dx\left(\frac{dy}{dx}\right)=\left(\frac{1}{cos(x+y)}\right)letx+y=tthen1+\left(\frac{dy}{dx}\right)=\left(\frac{dt}{dx}\right)or\left(\frac{dy}{dx}\right)=\left(\frac{dt}{dx}\right)-1\therefore \left(\frac{dt}{dx}\right)-1=\left(\frac{1}{cost}\right)\left(\frac{dt}{dx}\right)=\left(\frac{1}{cost}\right)+1\left(\frac{dt}{dx}\right)=\left(\frac{1+cost}{cost}\right)\int \left(\frac{costdt}{1+cost}\right)=\int dx\int \left(\frac{cost(1-cost)}{1^2-co{s}^{2}t}\right)dt=x+c\int \left(\frac{cost-co{s}^{2}t}{si{n}^{2}t}\right)dt=x+c\int cottcosectdt-\int co{t}^{2}tdt=x+c-cosect-\int (cose{c}^{2}t-1)dt=x+c-cosect+cott+t=x+c-cosec(x+y)+cot(x+y)+x+y=x+c-cosec(x+y)+cot(x+y)+y=c$