AP DSC School Assistant Math Model Test 3

Concept:The problem uses a trigonometric simplification followed by the property of definite integrals with substitution $x \to \frac{\pi}{2} - x$ to show that two integrals are equal, leading to cancellation.Explanation:Let $I = \displaystyle\int_{0}^{\frac{\pi}{2}} \frac{1-\cot x}{\operatorname{cosec} x+\cos x} \, dx$.Rewrite the integrand using basic identities: $\cot x = \frac{\cos x}{\sin x}$ and $\operatorname{cosec} x = \frac{1}{\sin x}$.Then numerator: $1 - \cot x = 1 - \frac{\cos x}{\sin x} = \frac{\sin x - \cos x}{\sin x}$.Denominator: $\operatorname{cosec} x + \cos x = \frac{1}{\sin x} + \cos x = \frac{1 + \sin x \cos x}{\sin x}$.Thus the integrand becomes $\displaystyle \frac{\frac{\sin x - \cos x}{\sin x}}{\frac{1 + \sin x \cos x}{\sin x}} = \frac{\sin x - \cos x}{1 + \sin x \cos x}$.So $I = \displaystyle\int_{0}^{\frac{\pi}{2}} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx$.Split the integral: $I = \displaystyle\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1 + \sin x \cos x} \, dx - \displaystyle\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1 + \sin x \cos x} \, dx$.Call $I_1 = \displaystyle\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1 + \sin x \cos x} \, dx$ and $I_2 = \displaystyle\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1 + \sin x \cos x} \, dx$.Now use the substitution $x \to \frac{\pi}{2} - x$ in $I_1$:$\sin\left(\frac{\pi}{2} - x\right) = \cos x$, $\cos\left(\frac{\pi}{2} - x\right) = \sin x$, and $\sin\left(\frac{\pi}{2} - x\right) \cos\left(\frac{\pi}{2} - x\right) = \cos x \sin x$, which is symmetric. Also the limits remain $0$ to $\frac{\pi}{2}$. Hence $I_1$ transforms exactly into $I_2$.Therefore $I_1 = I_2$, and $I = I_1 - I_2 = 0$.Answer:Option A: 0