MHTCET 15 Apr 2026 Shift 2 Paper

Concept:Use the substitution $u = x + y$ to reduce the equation to a separable form, then integrate using the half-angle substitution $t = \tan \frac{u}{2}$.Explanation:Let $u = x + y$.Differentiate with respect to $x$:$$\frac{du}{dx} = 1 + \frac{dy}{dx}$$Given $\frac{dy}{dx} = \sin(x+y) + \cos(x+y) = \sin u + \cos u$.Therefore,$$\frac{du}{dx} = 1 + \sin u + \cos u$$Separate the variables:$$\frac{du}{1 + \sin u + \cos u} = dx$$Integrate both sides:$$\int \frac{du}{1 + \sin u + \cos u} = \int dx$$Use the substitution $t = \tan \frac{u}{2}$.Then,$$\sin u = \frac{2t}{1+t^2}, \quad \cos u = \frac{1-t^2}{1+t^2}, \quad du = \frac{2\,dt}{1+t^2}$$Simplify the denominator:$$1 + \sin u + \cos u = 1 + \frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2} = \frac{2(1+t)}{1+t^2}$$Thus,$$\int \frac{du}{1 + \sin u + \cos u} = \int \frac{\frac{2\,dt}{1+t^2}}{\frac{2(1+t)}{1+t^2}} = \int \frac{dt}{1+t}$$Integrating gives:$$\log|1+t| = x + c$$Substitute back $t = \tan \frac{u}{2}$ and $u = x + y$:$$\log \left[1 + \tan \frac{x+y}{2}\right] = x + c$$This matches option C.Answer:$$\boxed{\log \left[1+\tan \left(\frac{x+y}{2}\right)\right]=x+c}$$Option C is correct.