### Related Formula King's Property of Definite Integrals: int_a^b f(x) dx = int_a^b f(a + b - x) dx$\int_a^b f(x) dx = \int_a^b f(a + b - x) dx$ ### Core Logic Let the given integral be: I = int_0^pi frac(x + 3)sin x1 + 3cos^2x dx quad dots (1)$I = \int_0^\pi \frac{(x + 3)\sin x}{1 + 3\cos^2x} dx \quad \dots (1)$ Applying King's property (x to pi - x$x \to \pi - x$): I = int_0^pi frac(pi - x + 3)sin(pi - x)1 + 3cos^2(pi - x) dx$I = \int_0^\pi \frac{(\pi - x + 3)\sin(\pi - x)}{1 + 3\cos^2(\pi - x)} dx$ I = int_0^pi frac(pi - x + 3)sin x1 + 3cos^2x dx quad dots (2)$I = \int_0^\pi \frac{(\pi - x + 3)\sin x}{1 + 3\cos^2x} dx \quad \dots (2)$ ### Step 1: Eliminate the x Variable Adding equations (1) and (2): 2I = int_0^pi frac[(x + 3) + (pi - x + 3)]sin x1 + 3cos^2x dx$2I = \int_0^\pi \frac{[(x + 3) + (\pi - x + 3)]\sin x}{1 + 3\cos^2x} dx$ 2I = (pi + 6)int_0^pi fracsin x1 + 3cos^2x dx$2I = (\pi + 6)\int_0^\pi \frac{\sin x}{1 + 3\cos^2x} dx$ Using the symmetric property int_0^2a f(x)dx = 2int_0^a f(x)dx$\int_0^{2a} f(x)dx = 2\int_0^a f(x)dx$ if f(2a-x)=f(x)$f(2a-x)=f(x)$: 2I = 2(pi + 6)int_0^pi/2 fracsin x1 + 3cos^2x dx$2I = 2(\pi + 6)\int_0^{\pi/2} \frac{\sin x}{1 + 3\cos^2x} dx$ I = (pi + 6)int_0^pi/2 fracsin x1 + 3cos^2x dx$I = (\pi + 6)\int_0^{\pi/2} \frac{\sin x}{1 + 3\cos^2x} dx$ ### Step 2: Solve Using Substitution Let t = sqrt3cos x$t = \sqrt{3}\cos x$. Then dt = -sqrt3sin x dx implies sin x dx = -fracdtsqrt3$dt = -\sqrt{3}\sin x dx \implies \sin x dx = -\frac{dt}{\sqrt{3}}$. Change in integration boundaries: - When x = 0 implies t = sqrt3$x = 0 \implies t = \sqrt{3}$ - When x = pi/2 implies t = 0$x = \pi/2 \implies t = 0$ Substituting into the integral: I = (pi + 6) int_sqrt3^0 frac-dt/sqrt31 + t^2 = fracpi + 6sqrt3 int_0^sqrt3 fracdt1 + t^2$I = (\pi + 6) \int_{\sqrt{3}}^0 \frac{-dt/\sqrt{3}}{1 + t^2} = \frac{\pi + 6}{\sqrt{3}} \int_0^{\sqrt{3}} \frac{dt}{1 + t^2}$ I = fracpi + 6sqrt3 left[ tan^-1t right]_0^sqrt3 = fracpi + 6sqrt3 left( tan^-1sqrt3 - 0 right)$I = \frac{\pi + 6}{\sqrt{3}} \left[ \tan^{-1}t \right]_0^{\sqrt{3}} = \frac{\pi + 6}{\sqrt{3}} \left( \tan^{-1}\sqrt{3} - 0 \right)$ I = fracpi + 6sqrt3 cdot fracpi3 = fracpi3sqrt3(pi + 6)$I = \frac{\pi + 6}{\sqrt{3}} \cdot \frac{\pi}{3} = \frac{\pi}{3\sqrt{3}}(\pi + 6)$ ### Pattern Recognition Whenever you encounter a linear x$x$ factor multiplying trigonometric components in a definite integral with symmetric limits like 0$0$ to pi$\pi$, executing King's property first is almost guaranteed to cleanly wipe out that variable element. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals

### Related Formula int_a^b f(x) \, dx = int_a^b f(a+b-x) \, dx$\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$ ### Core Logic Perform variable substitution to match the arguments and limit bounds across both separate integral functions before invoking King's property. ### Step 1: Perform Base Transformation Substitution In I_1$I_1$, let 2x = t implies 2dx = dt$2x = t \implies 2dx = dt$. Limits mapping: x = -1/2 implies t = -1$x = -1/2 \implies t = -1$; x = 1 implies t = 2$x = 1 \implies t = 2$. I_1 = frac12 int_-1^2 t f(t(1-t)) \, dt implies 2I_1 = int_-1^2 t f(t(1-t)) \, dt$I_1 = \frac{1}{2} \int_{-1}^{2} t f(t(1-t)) \, dt \implies 2I_1 = \int_{-1}^{2} t f(t(1-t)) \, dt$ ### Step 2: Invoke Integral Mirror Properties Apply the identity using parameters (a+b-t) = (1-t)$(a+b-t) = (1-t)$: 2I_1 = int_-1^2 (1-t) f((1-t)(1-(1-t))) \, dt$2I_1 = \int_{-1}^{2} (1-t) f((1-t)(1-(1-t))) \, dt$ 2I_1 = int_-1^2 f(t(1-t)) \, dt - int_-1^2 t f(t(1-t)) \, dt$2I_1 = \int_{-1}^{2} f(t(1-t)) \, dt - \int_{-1}^{2} t f(t(1-t)) \, dt$ ### Step 3: Final Matrix Matching Evaluation Notice component blocks align exactly with I_2$I_2$ definition values: 2I_1 = I_2 - 2I_1 implies 4I_1 = I_2$2I_1 = I_2 - 2I_1 \implies 4I_1 = I_2$ fracI_2I_1 = 4$\frac{I_2}{I_1} = 4$ ### Pattern Recognition Symmetric transformations highlighting factor expressions like x(1-x)$x(1-x)$ coupled with an external linear multiplier term x$x$ naturally simplify to half-weight area forms using reflection rules. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals

### Related Formula int x sin(pi x) \, dx = -fracxpicos(pi x) + fracsin(pi x)pi^2$\int x \sin(\pi x) \, dx = -\frac{x}{\pi}\cos(\pi x) + \frac{\sin(\pi x)}{\pi^2}$ ### Core Logic Track sign configurations across the target integration segments to drop absolute modulus walls effectively at clean quadrant intervals. ### Step 1: Break Apart the Modulus Domain For x in [-1, 1]$x \in [-1, 1]$, product value elements x sin(pi x) ge 0$x \sin(\pi x) \ge 0$. For x in [1, 3/2]$x \in [1, 3/2]$, values drop below zero: I = pi^2 left\ int_-1^1 x sin(pi x) \, dx - int_1^3/2 x sin(pi x) \, dx right\$I = \pi^2 \left\{ \int_{-1}^{1} x \sin(\pi x) \, dx - \int_{1}^{3/2} x \sin(\pi x) \, dx \right\}$ ### Step 2: Integrate the Even Function Block Since x sin(pi x)$x \sin(\pi x)$ is symmetric and even: int_-1^1 x sin(pi x) \, dx = 2 int_0^1 x sin(pi x) \, dx = 2 left[ -fracxpicos(pi x) + fracsin(pi x)pi^2 right]_0^1 = frac2pi$\int_{-1}^{1} x \sin(\pi x) \, dx = 2 \int_{0}^{1} x \sin(\pi x) \, dx = 2 \left[ -\frac{x}{\pi}\cos(\pi x) + \frac{\sin(\pi x)}{\pi^2} \right]_{0}^{1} = \frac{2}{\pi}$ ### Step 3: Subtract the Inverse Segment Evaluating boundary limits across the secondary phase track: int_1^3/2 x sin(pi x) \, dx = left[ -fracxpicos(pi x) + fracsin(pi x)pi^2 right]_1^3/2 = left( 0 - frac1pi^2 right) - left( frac1pi right) = -frac1pi^2 - frac1pi$\int_{1}^{3/2} x \sin(\pi x) \, dx = \left[ -\frac{x}{\pi}\cos(\pi x) + \frac{\sin(\pi x)}{\pi^2} \right]_{1}^{3/2} = \left( 0 - \frac{1}{\pi^2} \right) - \left( \frac{1}{\pi} \right) = -\frac{1}{\pi^2} - \frac{1}{\pi}$ I = pi^2 left\ frac2pi - left(-frac1pi^2 - frac1piright) right\ = pi^2 left( frac3pi + frac1pi^2 right) = 3pi + 1$I = \pi^2 \left\{ \frac{2}{\pi} - \left(-\frac{1}{\pi^2} - \frac{1}{\pi}\right) \right\} = \pi^2 \left( \frac{3}{\pi} + \frac{1}{\pi^2} \right) = 3\pi + 1$ ### Pattern Recognition Products of two odd tracking metrics (like linear variable x$x$ matched with sinusoidal waves) yield overall even systems, enabling rapid evaluation over center-aligned domains. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals

### Related Formula Additivity property of definite integrals over split interval boundary points: int_a^c g(x) \, dx = int_a^b g(x) \, dx + int_b^c g(x) \, dx$\int_{a}^{c} g(x) \, dx = \int_{a}^{b} g(x) \, dx + \int_{b}^{c} g(x) \, dx$ ### Core Logic Separate the integration tracking flow into two component operations: I_1 = int_0^fracpi4 sinleft|4x - fracpi12 ight| \, dx$I_1 = \int_{0}^{\frac{pi}{4}} \sin\left|4x - \frac{pi}{12} ight| \, dx$ I_2 = int_0^fracpi4 [2sin x] \, dx$I_2 = \int_{0}^{\frac{pi}{4}} [2\sin x] \, dx$ ### Step 1: Solve Modulus Integral Term The argument changes sign inside absolute value bounds when 4x - fracpi12 = 0 implies x = fracpi48$4x - \frac{pi}{12} = 0 \implies x = \frac{pi}{48}$: I_1 = int_0^fracpi48 -sinleft(4x - fracpi12 ight) \, dx + int_fracpi48^fracpi4 sinleft(4x - fracpi12 ight) \, dx$I_1 = \int_{0}^{\frac{pi}{48}} -\sin\left(4x - \frac{pi}{12} ight) \, dx + \int_{\frac{pi}{48}}^{\frac{pi}{4}} \sin\left(4x - \frac{pi}{12} ight) \, dx$ = frac14 left[ cosleft(4x - fracpi12 ight) right]_0^fracpi48 - frac14 left[ cosleft(4x - fracpi12 ight) right]_fracpi48^fracpi4$= \frac{1}{4} \left[ \cos\left(4x - \frac{pi}{12} ight) \right]_{0}^{\frac{pi}{48}} - \frac{1}{4} \left[ \cos\left(4x - \frac{pi}{12} ight) \right]_{\frac{pi}{48}}^{\frac{pi}{4}}$ Evaluating across numerical boundaries gives: 24 cdot I_1 = 24left(frac12right) = 12$24 \cdot I_1 = 24\left(\frac{1}{2}\right) = 12$ ### Step 2: Solve Greatest Integer Component and Combine Analyze step limits inside greatest integer block [2sin x]$[2\sin x]$: For x in left[0, fracpi6right)$x \in \left[0, \frac{pi}{6}\right)$, 0 le 2sin x < 1 implies [2sin x] = 0$0 \le 2\sin x < 1 \implies [2\sin x] = 0$. For x in left[fracpi6, fracpi4right]$x \in \left[\frac{pi}{6}, \frac{pi}{4}\right]$, 1 le 2sin x < sqrt2 implies [2sin x] = 1$1 \le 2\sin x < \sqrt{2} \implies [2\sin x] = 1$. I_2 = int_0^fracpi6 0 \, dx + int_fracpi6^fracpi4 1 \, dx = fracpi4 - fracpi6 = fracpi12$I_2 = \int_{0}^{\frac{pi}{6}} 0 \, dx + \int_{\frac{pi}{6}}^{\frac{pi}{4}} 1 \, dx = \frac{pi}{4} - \frac{pi}{6} = \frac{pi}{12}$ Combine both sections aggregated by multiplier 24: textTotal = 12 + 24left(fracpi12 ight) = 2pi + 12$\text{Total} = 12 + 24\left(\frac{pi}{12} ight) = 2\pi + 12$ Comparing directly with expression statement parameters 2pi + alpha$2\pi + \alpha$ isolates response value: alpha = 12$\alpha = 12$ ### Pattern Recognition Modulus arguments and step functions require isolating inflection transition numbers directly to break integrations up neatly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integration

### Related Formula King's property for definite integration: int_a^b f(x) dx = int_a^b f(a+b-x) dx$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ ### Core Logic Apply the identity x to -x$x \to -x$ to the integral: I = int_-fracpi2^fracpi2 frac96x^2 cos^2 x1 + e^x dx = int_-fracpi2^fracpi2 frac96x^2 cos^2 x1 + e^-x dx$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96x^2 \cos^2 x}{1 + e^x} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96x^2 \cos^2 x}{1 + e^{-x}} dx$ ### Step 1: Adding both integral variations Adding the equations eliminates the exponential denominator term (1+e^x$1+e^x$): 2I = int_-fracpi2^fracpi2 96x^2 cos^2 x cdot left[frac11+e^x + frace^x1+e^xright] dx$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 96x^2 \cos^2 x \cdot \left[\frac{1}{1+e^x} + \frac{e^x}{1+e^x}\right] dx$ I = 48 int_0^fracpi2 x^2 (1 + cos 2x) dx$I = 48 \int_{0}^{\frac{\pi}{2}} x^2 (1 + \cos 2x) dx$ ### Step 2: Evaluating the integrated components Integrating by parts gives: I = pi (2pi^2 - 12)$I = \pi (2\pi^2 - 12)$ Matching coefficients with the template: alpha = 2$\alpha = 2$ and \beta = -12. (alpha + beta)^2 = (2 - 12)^2 = (-10)^2 = 100$(\alpha + \beta)^2 = (2 - 12)^2 = (-10)^2 = 100$ ### Pattern Recognition Exponential denominators like 1+e^x$1+e^x$ in symmetric integral intervals are prime candidates for simplification using King's property. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Definite Integrals