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 King's Property for symmetric integration limits: int_-a^a f(x) \, dx = int_-a^a f(-x) \, dx$$\int_{-a}^{a} f(x) \, dx = \int_{-a}^{a} f(-x) \, dx$$ ### Core Logic Let the given definite integral be I$I$: I = int_-pi/2^pi/2frac8sqrt2cos x\,dx(1+e^sin x)(1+sin^4x) quad implies (1)$$I = \int_{-\pi/2}^{\pi/2}\frac{8\sqrt{2}\cos x\,dx}{(1+e^{\sin x})(1+\sin^{4}x)} \quad \implies (1)$$ Apply King's property by replacing x$x$ with -x$-x$ (since -pi/2 + pi/2 = 0$-\pi/2 + \pi/2 = 0$): I = int_-pi/2^pi/2frac8sqrt2cos(-x)\,dx(1+e^sin(-x))(1+sin^4(-x))$$I = \int_{-\pi/2}^{\pi/2}\frac{8\sqrt{2}\cos(-x)\,dx}{(1+e^{\sin(-x)})(1+\sin^{4}(-x))}$$ I = int_-pi/2^pi/2frac8sqrt2cos x\,dx(1+e^-sin x)(1+sin^4x) = int_-pi/2^pi/2frac8sqrt2cos x cdot e^sin x\,dx(e^sin x+1)(1+sin^4x) quad implies (2)$$I = \int_{-\pi/2}^{\pi/2}\frac{8\sqrt{2}\cos x\,dx}{(1+e^{-\sin x})(1+\sin^{4}x)} = \int_{-\pi/2}^{\pi/2}\frac{8\sqrt{2}\cos x \cdot e^{\sin x}\,dx}{(e^{\sin x}+1)(1+\sin^{4}x)} \quad \implies (2)$$ ### Step 1: Simplify by Adding Expressions Adding equations (1) and (2): 2I = int_-pi/2^pi/2 frac8sqrt2cos x(1 + e^sin x)\,dx(1+e^sin x)(1+sin^4x)$$2I = \int_{-\pi/2}^{\pi/2} \frac{8\sqrt{2}\cos x(1 + e^{\sin x})\,dx}{(1+e^{\sin x})(1+\sin^{4}x)}$$ 2I = int_-pi/2^pi/2 frac8sqrt2cos x\,dx1+sin^4x$$2I = \int_{-\pi/2}^{\pi/2} \frac{8\sqrt{2}\cos x\,dx}{1+\sin^{4}x}$$ Since the integrand is even, we can change the limits from 0$0$ to pi/2$\pi/2$: 2I = 2 int_0^pi/2 frac8sqrt2cos x\,dx1+sin^4x implies I = int_0^pi/2 frac8sqrt2cos x\,dx1+sin^4x$$2I = 2 \int_{0}^{\pi/2} \frac{8\sqrt{2}\cos x\,dx}{1+\sin^{4}x} \implies I = \int_{0}^{\pi/2} \frac{8\sqrt{2}\cos x\,dx}{1+\sin^{4}x}$$ ### Step 2: Substitution and Algebraic Deconstruction Let sin x = t implies cos x \, dx = dt$\sin x = t \implies \cos x \, dx = dt$. Limits change from 0$0$ to 1$1$: I = int_0^1 frac8sqrt2\,dt1+t^4 = 4sqrt2 int_0^1 frac2\,dt1+t^4$$I = \int_{0}^{1} \frac{8\sqrt{2}\,dt}{1+t^4} = 4\sqrt{2} \int_{0}^{1} \frac{2\,dt}{1+t^4}$$ Dividing the numerator and denominator by t^2$t^2$, we write it as two distinct expressions: I = 4sqrt2 left[ int_0^1 frac1+frac1t^2t^2+frac1t^2 \, dt - int_0^1 frac1-frac1t^2t^2+frac1t^2 \, dt right]$$I = 4\sqrt{2} \left[ \int_{0}^{1} \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}} \, dt - \int_{0}^{1} \frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}} \, dt \right]$$ I = 4sqrt2 left[ int_-infty^0 fracdzz^2+2 - int_infty^2 fracdkk^2-2 right]$$I = 4\sqrt{2} \left[ \int_{-\infty}^{0} \frac{dz}{z^2+2} - \int_{\infty}^{2} \frac{dk}{k^2-2} \right]$$ where z = t - frac1t$z = t - \frac{1}{t}$ and k = t + frac1t$k = t + \frac{1}{t}$. ### Step 3: Integrate and Evaluate Parameters Evaluating standard anti-derivatives: I = 4sqrt2 left[ frac1sqrt2tan^-1left(fraczsqrt2right) right]_-infty^0 - 4sqrt2 left[ frac12sqrt2lnleft|frack-sqrt2k+sqrt2right| right]_infty^2$$I = 4\sqrt{2} \left[ \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{z}{\sqrt{2}}\right) \right]_{-\infty}^{0} - 4\sqrt{2} \left[ \frac{1}{2\sqrt{2}}\ln\left|\frac{k-\sqrt{2}}{k+\sqrt{2}}\right| \right]_{\infty}^{2}$$ I = 4sqrt2left(fracpi2sqrt2right) - 2 lnleft|frac2-sqrt22+sqrt2right| = 2pi - 2ln(sqrt2-1)^2$$I = 4\sqrt{2}\left(\frac{\pi}{2\sqrt{2}}\right) - 2 \ln\left|\frac{2-\sqrt{2}}{2+\sqrt{2}}\right| = 2\pi - 2\ln(\sqrt{2}-1)^2$$ I = 2pi + 2ln(3+2sqrt2)$$I = 2\pi + 2\ln(3+2\sqrt{2})$$ Matching parameters yields alpha = 2$\alpha = 2$ and beta = 2$\beta = 2$. Thus: alpha^2 + beta^2 = 2^2 + 2^2 = 8$$\alpha^2 + \beta^2 = 2^2 + 2^2 = 8$$ ### Pattern Recognition Sees: Exponential variables causing asymmetry in symmetric integration bounds. Shortcut: Using King's rule completely eliminates the confusing e^sin x$e^{\sin x}$ factor, leaving behind a straightforward rational trigonometric configuration. ### 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 quad text(King's Rule)$$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx \quad \text{(King's Rule)}$$ ### Core Logic First, find the orthocentre (a,b)$(a,b)$ of Delta ABC$\Delta ABC$ with vertices A(1, 2)$A(1, 2)$, B(2, 3)$B(2, 3)$, and C(3, 1)$C(3, 1)$. Slope of AB = frac3-22-1 = 1$AB = \frac{3-2}{2-1} = 1$. The altitude from C$C$ onto AB$AB$ must be perpendicular to AB$AB$, so its slope is -1$-1$. Equation of altitude from C(3,1)$C(3,1)$: y - 1 = -1(x - 3) Rightarrow x + y = 4$$y - 1 = -1(x - 3) \Rightarrow x + y = 4$$ The orthocentre (a, b)$(a, b)$ lies on all altitudes, including this one. Thus, it satisfies a + b = 4$a + b = 4$. ### Step 1: Applying Definite Integral Properties Given I_1 = int_a^b x sin(4x-x^2) dx$I_1 = \int_a^b x \sin(4x-x^2) dx$, let's rewrite the argument of sine: 4x - x^2 = x(4-x)$$4x - x^2 = x(4-x)$$ Apply King's Rule replacing x$x$ with (a+b-x)$(a+b-x)$. Since we proved a+b = 4$a+b = 4$, substitute x$x$ with (4-x)$(4-x)$: I_1 = int_a^b (4-x) sin((4-x)(4 - (4-x))) dx$$I_1 = \int_a^b (4-x) \sin((4-x)(4 - (4-x))) dx$$ I_1 = int_a^b (4-x) sin((4-x)x) dx$$I_1 = \int_a^b (4-x) \sin((4-x)x) dx$$ I_1 = int_a^b (4-x) sin(4x-x^2) dx$$I_1 = \int_a^b (4-x) \sin(4x-x^2) dx$$ ### Step 2: Evaluating the Integral Ratio Expand the newly formed integral: I_1 = 4 int_a^b sin(4x-x^2) dx - int_a^b x sin(4x-x^2) dx$$I_1 = 4 \int_a^b \sin(4x-x^2) dx - \int_a^b x \sin(4x-x^2) dx$$ Notice that the second term is I_1$I_1$ and the first integral is I_2$I_2$: I_1 = 4I_2 - I_1 Rightarrow 2I_1 = 4I_2 Rightarrow fracI_1I_2 = 2$$I_1 = 4I_2 - I_1 \Rightarrow 2I_1 = 4I_2 \Rightarrow \frac{I_1}{I_2} = 2$$ ### Step 3: Final Output Evaluation We need the value of 36 fracI_1I_2$36 \frac{I_1}{I_2}$: 36 times 2 = 72$$36 \times 2 = 72$$ ### Pattern Recognition Whenever you see int_a^b x cdot f(x(a+b-x)) dx$\int_a^b x \cdot f(x(a+b-x)) dx$, immediately apply King's Rule to factor out x$x$. You rarely need the individual values of the integration bounds, only their sum. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Straight Lines Class 12 Maths: Definite Integration

### Related Formula int x^n dx = fracx^n+1n+1$$\int x^n dx = \frac{x^{n+1}}{n+1}$$ ### Core Logic To evaluate integrals with sum of square roots in the denominator, multiply and divide by the conjugate to rationalize it. I = int_0^1fracsqrt3+x-sqrt1+x(sqrt3+x+sqrt1+x)(sqrt3+x-sqrt1+x)dx$$I = \int_{0}^{1}\frac{\sqrt{3+x}-\sqrt{1+x}}{(\sqrt{3+x}+\sqrt{1+x})(\sqrt{3+x}-\sqrt{1+x})}dx$$ I = int_0^1fracsqrt3+x-sqrt1+x(3+x) - (1+x)dx$$I = \int_{0}^{1}\frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x) - (1+x)}dx$$ I = frac12 int_0^1 (sqrt3+x - sqrt1+x) dx$$I = \frac{1}{2} \int_{0}^{1} (\sqrt{3+x} - \sqrt{1+x}) dx$$ ### Step 1: Integration and Bounds Setup Integrate the resulting expression: I = frac12 left[ frac(3+x)^3/23/2 - frac(1+x)^3/23/2 right]_0^1$$I = \frac{1}{2} \left[ \frac{(3+x)^{3/2}}{3/2} - \frac{(1+x)^{3/2}}{3/2} \right]_{0}^{1}$$ I = frac12 cdot frac23 left[ (3+x)^3/2 - (1+x)^3/2 right]_0^1$$I = \frac{1}{2} \cdot \frac{2}{3} \left[ (3+x)^{3/2} - (1+x)^{3/2} \right]_{0}^{1}$$ I = frac13 left[ ((4)^3/2 - (2)^3/2) - ((3)^3/2 - (1)^3/2) right]$$I = \frac{1}{3} \left[ ((4)^{3/2} - (2)^{3/2}) - ((3)^{3/2} - (1)^{3/2}) \right]$$ ### Step 2: Term Simplification Evaluate the boundary powers: 4^3/2 = 8$4^{3/2} = 8$ 2^3/2 = 2sqrt2$2^{3/2} = 2\sqrt{2}$ 3^3/2 = 3sqrt3$3^{3/2} = 3\sqrt{3}$ 1^3/2 = 1$1^{3/2} = 1$ Substitute back into the expression: I = frac13 [ 8 - 2sqrt2 - 3sqrt3 + 1 ] = frac13 [ 9 - 2sqrt2 - 3sqrt3 ]$$I = \frac{1}{3} [ 8 - 2\sqrt{2} - 3\sqrt{3} + 1 ] = \frac{1}{3} [ 9 - 2\sqrt{2} - 3\sqrt{3} ]$$ I = 3 - frac23sqrt2 - sqrt3$$I = 3 - \frac{2}{3}\sqrt{2} - \sqrt{3}$$ ### Step 3: Finding Co-efficients Comparing with a+bsqrt2+csqrt3$a+b\sqrt{2}+c\sqrt{3}$ yields: a = 3$a = 3$, b = -frac23$b = -\frac{2}{3}$, c = -1$c = -1$ Compute 2a+3b-4c$2a+3b-4c$: 2(3) + 3left(-frac23right) - 4(-1)$$2(3) + 3\left(-\frac{2}{3}\right) - 4(-1)$$ 6 - 2 + 4 = 8$6 - 2 + 4 = 8$ ### Pattern Recognition Whenever you see a sum of square roots in the denominator of an integrand, the immediate algorithmic next step is rationalization. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Definite Integration