MHTCET 16 Apr 2026 Shift 1 Paper

Concept:Simplify the integrand using the identity $\cos^2 x = 1 - \sin^2 x$, then use substitution $u = \sin x$.Explanation:Write $\cos^3 x = \cos x \cdot \cos^2 x$.Use $\cos^2 x = 1 - \sin^2 x$, so the numerator becomes $(1 - \sin^2 x)\cos x$.The denominator is $\sin^2 x + \sin x = \sin x(\sin x + 1)$.So the integral is:$$\int \frac{(1-\sin^2 x)\cos x}{\sin x(\sin x+1)}\,dx$$Factor the numerator: $1 - \sin^2 x = (1-\sin x)(1+\sin x)$.Cancel $(1+\sin x)$ from numerator and denominator:$$\int \frac{1-\sin x}{\sin x}\cos x\,dx$$Let $u = \sin x$, then $du = \cos x\,dx$.The integral becomes:$$\int \frac{1-u}{u}\,du = \int\left(\frac{1}{u}-1\right)du$$This evaluates to $\log|u| - u + C$.Substitute back $u = \sin x$:$$\log|\sin x| - \sin x + C$$This matches Option B.Answer:Option B: $\log(\sin x) - \sin x + c$, where $c$ is the constant of integration.