MHTCET 23 Apr 2025 Shift 2 Paper

Concept:Use the substitution $\sin\theta=t$ to convert the integral into a simple rational form, then apply partial fractions.Explanation:Let $I=\int \frac{(5\sin\theta-2)\cos\theta}{5-\cos^2\theta-4\sin\theta}\,d\theta$.Put $t=\sin\theta$, so $\cos\theta\,d\theta=dt$.Since $\cos^2\theta=1-t^2$, the denominator becomes:$5-(1-t^2)-4t=t^2-4t+4=(t-2)^2$.Therefore $I=\int \frac{5t-2}{(t-2)^2}\,dt$.Write $\frac{5t-2}{(t-2)^2}=\frac{A}{t-2}+\frac{B}{(t-2)^2}$.Multiplying by $(t-2)^2$ gives $5t-2=A(t-2)+B$.Comparing coefficients, we get $A=5$ and $B=8$.So $I=5\int \frac{dt}{t-2}+8\int \frac{dt}{(t-2)^2}$.This gives $I=5\log|t-2|-\frac{8}{t-2}+c$.Replacing $t=\sin\theta$,$I=5\log|\sin\theta-2|-\frac{8}{\sin\theta-2}+c$.Since $\sin\theta-2<0$, this can also be written as $5\log(2-\sin\theta)-\frac{8}{\sin\theta-2}+c$.Answer:$\int \frac{(5\sin\theta-2)\cos\theta}{5-\cos^2\theta-4\sin\theta}\,d\theta=5\log|\sin\theta-2|-\frac{8}{\sin\theta-2}+c$, where $c$ is the constant of integration.