### Related Formula int fracdta^2 - b^2 t^2 = frac12a ln left| fraca+bta-bt right|$\int \frac{dt}{a^2 - b^2 t^2} = \frac{1}{2a} \ln \left| \frac{a+bt}{a-bt} \right|$ sin 2theta = 1 - (sintheta - costheta)^2$\sin 2\theta = 1 - (\sin\theta - \cos\theta)^2$ ### Core Logic Let sintheta - costheta = t$\sin\theta - \cos\theta = t$. Then (costheta + sintheta)dtheta = dt$(\cos\theta + \sin\theta)d\theta = dt$. Transform limits: When theta = 0 implies t = 0 - 1 = -1$\theta = 0 \implies t = 0 - 1 = -1$ When theta = fracpi4 implies t = frac1sqrt2 - frac1sqrt2 = 0$\theta = \frac{\pi}{4} \implies t = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$ ### Step 1: Perform the algebraic substitution Express the denominator base: 9 + 16sin 2theta = 9 + 16[1 - t^2] = 25 - 16t^2$9 + 16\sin 2\theta = 9 + 16[1 - t^2] = 25 - 16t^2$ The integral transforms to: I = 80 int_-1^0 fracdt25 - 16t^2 = frac8016 int_-1^0 fracdtleft(frac54right)^2 - t^2$I = 80 \int_{-1}^{0} \frac{dt}{25 - 16t^2} = \frac{80}{16} \int_{-1}^{0} \frac{dt}{\left(\frac{5}{4}\right)^2 - t^2}$ ### Step 2: Execute Integral Calculation I = 5 left[ frac12left(frac54right) ln left| fracfrac54 + tfrac54 - t right| right]_-1^0$I = 5 \left[ \frac{1}{2\left(\frac{5}{4}\right)} \ln \left| \frac{\frac{5}{4} + t}{\frac{5}{4} - t} \right| \right]_{-1}^{0}$ I = 2 left[ ln(1) - lnleft( frac1/49/4 right) right] = 2 left[ 0 - lnleft(frac19right) right] = 2ln(9) = 4ln(3)$I = 2 \left[ \ln(1) - \ln\left( \frac{1/4}{9/4} \right) \right] = 2 \left[ 0 - \ln\left(\frac{1}{9}\right) \right] = 2\ln(9) = 4\ln(3)$ ### Pattern Recognition Whenever (sintheta + costheta)$(\sin\theta + \cos\theta)$ sits inside the numerator, instantly choose t = sintheta - costheta$t = \sin\theta - \cos\theta$ as your core linear substitution engine to clean denominators. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals

### Related Formula textLeibniz Integral Rule for differentiation: fracddxint_0^x f(t) \, dt = f(x)$\text{Leibniz Integral Rule for differentiation: } \frac{d}{dx}\int_{0}^{x} f(t) \, dt = f(x)$ ### Core Logic Perform variable substitution inside the integral: let lambda x = t implies dlambda = frac1x dt$\lambda x = t \implies d\lambda = \frac{1}{x} dt$. When lambda = 0 implies t = 0$\lambda = 0 \implies t = 0$; when lambda = 1 implies t = x$\lambda = 1 \implies t = x$. The equation transforms to: frac1x int_0^x f(t) \, dt = a f(x) implies int_0^x f(t) \, dt = a x f(x)$\frac{1}{x} \int_{0}^{x} f(t) \, dt = a f(x) \implies \int_{0}^{x} f(t) \, dt = a x f(x)$ ### Step 1: Differentiate with respect to x Using Leibniz rule and product rule: f(x) = a [x f'(x) + f(x)]$f(x) = a [x f'(x) + f(x)]$ (1 - a)f(x) = a x f'(x) implies fracf'(x)f(x) = frac1-aa frac1x$(1 - a)f(x) = a x f'(x) \implies \frac{f'(x)}{f(x)} = \frac{1-a}{a} \frac{1}{x}$ Integrating both sides yields: ln f(x) = left(frac1-aaright)ln x + c implies f(x) = C x^frac1-aa$\ln f(x) = \left(\frac{1-a}{a}\right)\ln x + c \implies f(x) = C x^{\frac{1-a}{a}}$ ### Step 2: Calculate Constants using boundaries Given f(1) = 1 implies C = 1$f(1) = 1 \implies C = 1$. Given f(16) = frac18 implies frac18 = (16)^frac1-aa implies 2^-3 = (2^4)^frac1-aa$f(16) = \frac{1}{8} \implies \frac{1}{8} = (16)^{\frac{1-a}{a}} \implies 2^{-3} = (2^4)^{\frac{1-a}{a}}$ -3 = frac4(1-a)a implies -3a = 4 - 4a implies a = 4$-3 = \frac{4(1-a)}{a} \implies -3a = 4 - 4a \implies a = 4$ Therefore, power exponent = frac1-44 = -frac34 implies f(x) = x^-frac34$= \frac{1-4}{4} = -\frac{3}{4} \implies f(x) = x^{-\frac{3}{4}}$. ### Step 3: Evaluate target derivative value Find the derivative: f'(x) = -frac34 x^-frac74$f'(x) = -\frac{3}{4} x^{-\frac{7}{4}}$ Substitute x = frac116$x = \frac{1}{16}$: f'left(frac116right) = -frac34 left(2^-4right)^-frac74 = -frac34 left(2^7right) = -frac34 times 128 = -96$f'\left(\frac{1}{16}\right) = -\frac{3}{4} \left(2^{-4}\right)^{-\frac{7}{4}} = -\frac{3}{4} \left(2^7\right) = -\frac{3}{4} \times 128 = -96$ Final requested computation calculation: 16 - f'left(frac116right) = 16 - (-96) = 112$16 - f'\left(\frac{1}{16}\right) = 16 - (-96) = 112$ ### Pattern Recognition Scaling inputs inside functional definite integrals tracks closely to homogenous Euler equation properties. Converting integrations quickly to local power functions reduces processing parameters. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Differential Equations

### 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$I$: I = int_0^fracpi4fracx \, dxsin^4(2x)+cos^4(2x)$I = \int_{0}^{\frac{\pi}{4}}\frac{x \, dx}{\sin^{4}(2x)+\cos^{4}(2x)}$ Substitute 2x = t implies 2dx = dt implies dx = frac12dt$2x = t \implies 2dx = dt \implies dx = \frac{1}{2}dt$. When x = 0 implies t = 0$x = 0 \implies t = 0$, and when x = fracpi4 implies t = fracpi2$x = \frac{\pi}{4} \implies t = \frac{\pi}{2}$. I = frac14 int_0^fracpi2 fract \, dtsin^4t + cos^4t quad implies (1) $I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{t \, dt}{\sin^{4}t + \cos^{4}t} \quad \implies (1) $ ### Step 1: Apply King's Property Applying the property int_0^a f(t) \, dt = int_0^a f(a-t) \, dt$\int_{0}^{a} f(t) \, dt = \int_{0}^{a} f(a-t) \, dt$: I = frac14 int_0^fracpi2 fracleft(fracpi2 - tright) dtsin^4left(fracpi2-tright) + cos^4left(fracpi2-tright)$I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2} - t\right) dt}{\sin^{4}\left(\frac{\pi}{2}-t\right) + \cos^{4}\left(\frac{\pi}{2}-t\right)}$ I = frac14 int_0^fracpi2 fracleft(fracpi2 - tright) dtcos^4t + sin^4t quad implies (2) $I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2} - t\right) dt}{\cos^{4}t + \sin^{4}t} \quad \implies (2) $ Adding equations (1) and (2): 2I = frac14 int_0^fracpi2 fracfracpi2 dtsin^4t + cos^4t$2I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{\frac{\pi}{2} dt}{\sin^{4}t + \cos^{4}t}$ 2I = fracpi8 int_0^fracpi2 fracdtsin^4t + cos^4t $2I = \frac{\pi}{8} \int_{0}^{\frac{\pi}{2}} \frac{dt}{\sin^{4}t + \cos^{4}t} $ 2I = fracpi8 int_0^fracpi2 fracsec^4t \, dttan^4t + 1 $2I = \frac{\pi}{8} \int_{0}^{\frac{\pi}{2}} \frac{\sec^{4}t \, dt}{\tan^{4}t + 1} $ 2I = fracpi8 int_0^fracpi2 frac(1 + tan^2t)sec^2t \, dttan^4t + 1$2I = \frac{\pi}{8} \int_{0}^{\frac{\pi}{2}} \frac{(1 + \tan^{2}t)\sec^{2}t \, dt}{\tan^{4}t + 1}$ ### Step 2: Substitution and Algebraic Limits Let tan t = y implies sec^2t \, dt = dy$\tan t = y \implies \sec^{2}t \, dt = dy$. When t = 0 implies y = 0$t = 0 \implies y = 0$, and when t = fracpi2 implies y = infty$t = \frac{\pi}{2} \implies y = \infty$. 2I = fracpi8 int_0^infty frac(1 + y^2) \, dy1 + y^4 $2I = \frac{\pi}{8} \int_{0}^{\infty} \frac{(1 + y^{2}) \, dy}{1 + y^{4}} $ I = fracpi16 int_0^infty frac1 + frac1y^2y^2 + frac1y^2 \, dy $I = \frac{\pi}{16} \int_{0}^{\infty} \frac{1 + \frac{1}{y^{2}}}{y^{2} + \frac{1}{y^{2}}} \, dy $ Now put y - frac1y = p implies left(1 + frac1y^2right) dy = dp$y - \frac{1}{y} = p \implies \left(1 + \frac{1}{y^{2}}\right) dy = dp$. When y to 0^+ implies p to -infty$y \to 0^+ \implies p \to -\infty$, and when y to infty implies p to infty$y \to \infty \implies p \to \infty$. Also, y^2 + frac1y^2 = p^2 + 2 = p^2 + (sqrt2)^2$y^{2} + \frac{1}{y^{2}} = p^{2} + 2 = p^{2} + (\sqrt{2})^{2}$. I = fracpi16 int_-infty^infty fracdpp^2 + (sqrt2)^2 $I = \frac{\pi}{16} \int_{-\infty}^{\infty} \frac{dp}{p^{2} + (\sqrt{2})^{2}} $ I = fracpi16sqrt2 left[ tan^-1left(fracpsqrt2right) right]_-infty^infty $I = \frac{\pi}{16\sqrt{2}} \left[ \tan^{-1}\left(\frac{p}{\sqrt{2}}\right) \right]_{-\infty}^{\infty} $ $I = fracpi16sqrt2 left( fracpi2 - left(-fracpi2right) right) = fracpi^216sqrt2 = fracsqrt2pi^232 $I = \frac{\pi}{16\sqrt{2}} \left( \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \right) = \frac{\pi^{2}}{16\sqrt{2}} = \frac{\sqrt{2}\pi^{2}}{32} $ ### Pattern Recognition Sees: Integrand containing x$x$ in the numerator and symmetric trigonometric functions in the denominator. Shortcut: The elimination of x$x$ using King's property is standard. For integrals containing frac1+y^21+y^4$\frac{1+y^2}{1+y^4}$, dividing by y^2$y^2$ transforms the denominator into a perfect square form left(y-frac1yright)^2+2$\left(y-\frac{1}{y}\right)^2+2$, making substitution trivial. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals Class 11 Mathematics: Trigonometric Identities

### 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