The integral int_0^pi frac(x + 3)sin x1 + 3cos^2x dx$\int_0^\pi \frac{(x + 3)\sin x}{1 + 3\cos^2x} dx$ is equal to:

### 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

### Core Logic We have two functional equations to solve before integrating. Equation 1: f(x) + 2fleft(frac1xright) = x^2 + 5$f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5$ Replace x$x$ with frac1x$\frac{1}{x}$: fleft(frac1xright) + 2f(x) = frac1x^2 + 5$$f\left(\frac{1}{x}\right) + 2f(x) = \frac{1}{x^2} + 5$$ Multiplying this new equation by 2$2$ and subtracting the original Equation 1 eliminates the fleft(frac1xright)$f\left(\frac{1}{x}\right)$ term: 4f(x) + 2fleft(frac1xright) - left(f(x) + 2fleft(frac1xright)right) = 2left(frac1x^2 + 5right) - (x^2 + 5)$$4f(x) + 2f\left(\frac{1}{x}\right) - \left(f(x) + 2f\left(\frac{1}{x}\right)\right) = 2\left(\frac{1}{x^2} + 5\right) - (x^2 + 5)$$ 3f(x) = frac2x^2 - x^2 + 5 implies f(x) = frac23x^2 - fracx^23 + frac53$$3f(x) = \frac{2}{x^2} - x^2 + 5 \implies f(x) = \frac{2}{3x^2} - \frac{x^2}{3} + \frac{5}{3}$$ ### Step 1: Finding alpha Integrate f(x)$f(x)$ from 1$1$ to 2$2$: alpha = int_1^2 left( frac23x^2 - fracx^23 + frac53 right) dx = left[ -frac23x - fracx^39 + frac5x3 right]_1^2$$\alpha = \int_{1}^{2} \left( \frac{2}{3x^2} - \frac{x^2}{3} + \frac{5}{3} \right) dx = \left[ -\frac{2}{3x} - \frac{x^3}{9} + \frac{5x}{3} \right]_{1}^{2}$$ alpha = left( -frac13 - frac89 + frac103 right) - left( -frac23 - frac19 + frac53 right) = frac199 - frac89 = frac119$$\alpha = \left( -\frac{1}{3} - \frac{8}{9} + \frac{10}{3} \right) - \left( -\frac{2}{3} - \frac{1}{9} + \frac{5}{3} \right) = \frac{19}{9} - \frac{8}{9} = \frac{11}{9}$$ Thus, 9alpha = 11$9\alpha = 11$. ### Step 2: Solving for g(x) and finding beta We are given 2g(x) - 3gleft(frac12right) = x$2g(x) - 3g\left(\frac{1}{2}\right) = x$. Substitute x = frac12$x = \frac{1}{2}$: 2gleft(frac12right) - 3gleft(frac12right) = frac12 implies -gleft(frac12right) = frac12 implies gleft(frac12right) = -frac12$$2g\left(\frac{1}{2}\right) - 3g\left(\frac{1}{2}\right) = \frac{1}{2} \implies -g\left(\frac{1}{2}\right) = \frac{1}{2} \implies g\left(\frac{1}{2}\right) = -\frac{1}{2}$$ Substitute this constant value back into the original equation: 2g(x) - 3left(-frac12right) = x implies 2g(x) + frac32 = x implies g(x) = fracx2 - frac34$$2g(x) - 3\left(-\frac{1}{2}\right) = x \implies 2g(x) + \frac{3}{2} = x \implies g(x) = \frac{x}{2} - \frac{3}{4}$$ Now find beta$\beta$: beta = int_1^2 left( fracx2 - frac34 right) dx = left[ fracx^24 - frac3x4 right]_1^2 = left( 1 - frac32 right) - left( frac14 - frac34 right) = -frac12 - left(-frac12right) = 0$$\beta = \int_{1}^{2} \left( \frac{x}{2} - \frac{3}{4} \right) dx = \left[ \frac{x^2}{4} - \frac{3x}{4} \right]_{1}^{2} = \left( 1 - \frac{3}{2} \right) - \left( \frac{1}{4} - \frac{3}{4} \right) = -\frac{1}{2} - \left(-\frac{1}{2}\right) = 0$$ ### Step 3: Calculating 9alpha + beta Combining our values: 9alpha + beta = 11 + 0 = 11$$9\alpha + \beta = 11 + 0 = 11$$ ### Pattern Recognition Functional equations involving x to frac1x$x \to \frac{1}{x}$ are easily solved by treating the swapped forms as a system of linear equations, allowing direct isolation of the underlying function. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Functional Equations

### Related Formula King's property of definite integrals: int_a^b f(x)mathrmdx = int_a^b f(a+b-x)mathrmdx$$\int_{a}^{b} f(x)\mathrm{d}x = \int_{a}^{b} f(a+b-x)\mathrm{d}x$$ ### Core Logic Let the given integral be I$I$. Apply King's property by substituting x to -x$x \to -x$: I = int_-1^1fracleft(1 + sqrt|x| + xright)e^-x + left(sqrt|x| + xright)e^xe^-x + e^x\,mathrmdx$$I = \int_{-1}^{1}\frac{\left(1 + \sqrt{|x| + x}\right)e^{-x} + \left(\sqrt{|x| + x}\right)e^{x}}{e^{-x} + e^{x}}\,\mathrm{d}x$$ Add both integral expressions 2I = I + I$2I = I + I$: 2I = int_-1^1 frac(e^x + e^-x) + left(sqrt|x| - x + sqrt|x| + xright)(e^x + e^-x)e^x + e^-x\,mathrmdx$$2I = \int_{-1}^{1} \frac{(e^x + e^{-x}) + \left(\sqrt{|x| - x} + \sqrt{|x| + x}\right)(e^x + e^{-x})}{e^x + e^{-x}}\,\mathrm{d}x$$ 2I = int_-1^1 left(1 + sqrt|x| - x + sqrt|x| + xright)\,mathrmdx$$2I = \int_{-1}^{1} \left(1 + \sqrt{|x| - x} + \sqrt{|x| + x}\right)\,\mathrm{d}x$$ ### Step 1: Apply Symmetry Properties The integrand is completely even. Hence, convert intervals: 2I = 2int_0^1 left(1 + sqrt|x| - x + sqrt|x| + xright)\,mathrmdx$$2I = 2\int_{0}^{1} \left(1 + \sqrt{|x| - x} + \sqrt{|x| + x}\right)\,\mathrm{d}x$$ For x in [0,1]$x \in [0,1]$, |x| = x implies sqrt|x| - x = 0$|x| = x \implies \sqrt{|x| - x} = 0$ and sqrt|x| + x = sqrt2x$\sqrt{|x| + x} = \sqrt{2x}$: I = int_0^1 (1 + sqrt2x)\,mathrmdx$$I = \int_{0}^{1} (1 + \sqrt{2x})\,\mathrm{d}x$$ ### Step 2: Final Integration Execution I = left[ x + sqrt2 cdot fracx^3/23/2 right]_0^1 = left[ x + frac2sqrt23x^3/2 right]_0^1$$I = \left[ x + \sqrt{2} \cdot \frac{x^{3/2}}{3/2} \right]_{0}^{1} = \left[ x + \frac{2\sqrt{2}}{3}x^{3/2} \right]_{0}^{1}$$ I = 1 + frac2sqrt23$$I = 1 + \frac{2\sqrt{2}}{3}$$ ### Pattern Recognition When functions involve combinations of exponential components (e^x, e^-x$e^x, e^{-x}$) over symmetric boundaries, adding the variable reflection eliminates exponential fractions instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integration

### Related Formula The beta function integral format satisfies: I(m,n) = int_0^1 x^m-1 (1-x)^n-1 dx$$I(m,n) = \int_{0}^{1} x^{m-1} (1-x)^{n-1} dx$$ ### Core Logic Let's combine the terms of the requested sum directly by inserting their respective definitions: I(9,14) = int_0^1 x^9-1 (1-x)^14-1 dx = int_0^1 x^8 (1-x)^13 dx$$I(9,14) = \int_{0}^{1} x^{9-1} (1-x)^{14-1} dx = \int_{0}^{1} x^8 (1-x)^{13} dx$$ I(10,13) = int_0^1 x^10-1 (1-x)^13-1 dx = int_0^1 x^9 (1-x)^12 dx$$I(10,13) = \int_{0}^{1} x^{10-1} (1-x)^{13-1} dx = \int_{0}^{1} x^9 (1-x)^{12} dx$$ ### Step 1: Factoring out common algebraic terms Summing the two components: I(9,14) + I(10,13) = int_0^1 left[ x^8 (1-x)^13 + x^9 (1-x)^12 right] dx$$I(9,14) + I(10,13) = \int_{0}^{1} \left[ x^8 (1-x)^{13} + x^9 (1-x)^{12} \right] dx$$ Factor out the common term x^8 (1-x)^12$x^8 (1-x)^{12}$ inside the integrand: = int_0^1 x^8 (1-x)^12 left[ (1-x) + x right] dx$$= \int_{0}^{1} x^8 (1-x)^{12} \left[ (1-x) + x \right] dx$$ = int_0^1 x^8 (1-x)^12 (1) dx$$= \int_{0}^{1} x^8 (1-x)^{12} (1) dx$$ = int_0^1 x^9-1 (1-x)^13-1 dx = I(9,13)$$= \int_{0}^{1} x^{9-1} (1-x)^{13-1} dx = I(9,13)$$ ### Pattern Recognition When dealing with linear combinations of beta functions with shifting parameter indices, directly writing down the definite integral expression often results in immediate algebraic cancellation or simplification via basic factoring. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals