Let f(t) = int left( frac1 - sin(log_e t)1 - cos(log_e t) right) dt, t > 1$f(t) = \int \left( \frac{1 - \sin(\log_e t)}{1 - \cos(\log_e t)} \right) dt, t > 1$. If f(e^pi/2) = -e^pi/2$f(e^{\pi/2}) = -e^{\pi/2}$ and f(e^pi/4) = alpha e^pi/4$f(e^{\pi/4}) = \alpha e^{\pi/4}$, then alpha$\alpha$ equals

### Related Formula textWhen powers of sine and cosine add up to a negative even integer, extract sec^2 x text and substitute tan x = t$$\text{When powers of sine and cosine add up to a negative even integer, extract } \sec^2 x \text{ and substitute } \tan x = t$$ ### Core Logic Integral: I = int sin^-11/2 x cos^-5/2 x \, dx$I = \int \sin^{-11/2} x \cos^{-5/2} x \, dx$ The sum of powers is -frac112 - frac52 = -8$-\frac{11}{2} - \frac{5}{2} = -8$. Convert the integrand entirely into terms of tan x$\tan x$ and sec x$\sec x$ by dividing and multiplying by cos^-11/2 x$\cos^{-11/2} x$. I = int left(fracsin xcos xright)^-11/2 cos^-11/2 x cos^-5/2 x \, dx$$I = \int \left(\frac{\sin x}{\cos x}\right)^{-11/2} \cos^{-11/2} x \cos^{-5/2} x \, dx$$ I = int (tan x)^-11/2 (cos x)^-8 \, dx$$I = \int (\tan x)^{-11/2} (\cos x)^{-8} \, dx$$ I = int (tan x)^-11/2 sec^8 x \, dx$$I = \int (\tan x)^{-11/2} \sec^8 x \, dx$$ ### Step 1: Integration by Substitution Rewrite sec^8 x = (sec^2 x)^3 sec^2 x = (1 + tan^2 x)^3 sec^2 x$\sec^8 x = (\sec^2 x)^3 \sec^2 x = (1 + \tan^2 x)^3 \sec^2 x$. I = int (tan x)^-11/2 (1 + tan^2 x)^3 sec^2 x \, dx$$I = \int (\tan x)^{-11/2} (1 + \tan^2 x)^3 \sec^2 x \, dx$$ Substitute t = tan x implies dt = sec^2 x \, dx$t = \tan x \implies dt = \sec^2 x \, dx$. I = int t^-11/2 (1 + t^2)^3 \, dt$$I = \int t^{-11/2} (1 + t^2)^3 \, dt$$ Expand (1 + t^2)^3 = 1 + 3t^2 + 3t^4 + t^6$(1 + t^2)^3 = 1 + 3t^2 + 3t^4 + t^6$. I = int t^-11/2 (1 + 3t^2 + 3t^4 + t^6) \, dt$$I = \int t^{-11/2} (1 + 3t^2 + 3t^4 + t^6) \, dt$$ I = int (t^-11/2 + 3t^-7/2 + 3t^-3/2 + t^1/2) \, dt$$I = \int (t^{-11/2} + 3t^{-7/2} + 3t^{-3/2} + t^{1/2}) \, dt$$ ### Step 2: Evaluating the Anti-derivatives Integrate term by term: I = fract^-9/2-9/2 + 3fract^-5/2-5/2 + 3fract^-1/2-1/2 + fract^3/23/2 + C$$I = \frac{t^{-9/2}}{-9/2} + 3\frac{t^{-5/2}}{-5/2} + 3\frac{t^{-1/2}}{-1/2} + \frac{t^{3/2}}{3/2} + C$$ I = -frac29t^-9/2 - frac65t^-5/2 - 6t^-1/2 + frac23t^3/2 + C$$I = -\frac{2}{9}t^{-9/2} - \frac{6}{5}t^{-5/2} - 6t^{-1/2} + \frac{2}{3}t^{3/2} + C$$ Substitute back t = tan x = frac1cot x$t = \tan x = \frac{1}{\cot x}$, which implies t^-a = (cot x)^a$t^{-a} = (\cot x)^a$. I = -frac29(cot x)^9/2 - frac65(cot x)^5/2 - frac61(cot x)^1/2 + frac23(cot x)^-3/2 + C$$I = -\frac{2}{9}(\cot x)^{9/2} - \frac{6}{5}(\cot x)^{5/2} - \frac{6}{1}(\cot x)^{1/2} + \frac{2}{3}(\cot x)^{-3/2} + C$$ ### Step 3: Variable Assignment Comparing with -fracp_1q_1(cot x)^9/2 - fracp_2q_2(cot x)^5/2 - fracp_3q_3(cot x)^1/2 + fracp_4q_4(cot x)^-3/2$-\frac{p_1}{q_1}(\cot x)^{9/2} - \frac{p_2}{q_2}(\cot x)^{5/2} - \frac{p_3}{q_3}(\cot x)^{1/2} + \frac{p_4}{q_4}(\cot x)^{-3/2}$: p_1 = 2, q_1 = 9$p_1 = 2, q_1 = 9$ p_2 = 6, q_2 = 5$p_2 = 6, q_2 = 5$ p_3 = 6, q_3 = 1$p_3 = 6, q_3 = 1$ p_4 = 2, q_4 = 3$p_4 = 2, q_4 = 3$ Calculate frac15 p_1 p_2 p_3 p_4q_1 q_2 q_3 q_4$\frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4}$: = frac15(2)(6)(6)(2)(9)(5)(1)(3) = frac15 times 144135 = frac2160135 = 16$$= \frac{15(2)(6)(6)(2)}{(9)(5)(1)(3)} = \frac{15 \times 144}{135} = \frac{2160}{135} = 16$$ ### Pattern Recognition If int sin^m x cos^n x \,dx$\int \sin^m x \cos^n x \,dx$ has m+n$m+n$ as a negative even integer, unconditionally extract sec^|m+n|x$\sec^{|m+n|}x$ and set tan x = t$\tan x = t$. The expansion expands gracefully into polynomial power rules without any trig substitution hassle. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Indefinite Integration

### Related Formula Area enclosed between curves is computed by breaking into integration regions at intersection points. ### Core Logic

Find intersection of 4x^2 + y^2 = 8$4x^2 + y^2 = 8$ and y^2 = 4x$y^2 = 4x$: 4x^2 + 4x - 8 = 0 implies x^2 + x - 2 = 0 implies x = 1 quad (x > 0)$$4x^2 + 4x - 8 = 0 \implies x^2 + x - 2 = 0 \implies x = 1 \quad (x > 0)$$ For x in [0, 1]$x \in [0, 1]$, region bounded by parabola y = pm 2sqrtx$y = \pm 2\sqrt{x}$. For x in [1, sqrt2]$x \in [1, \sqrt{2}]$, region bounded by ellipse y = pm sqrt8 - 4x^2$y = \pm \sqrt{8 - 4x^2}$. ### Step 1: Definite Integration textArea = 2 int_0^1 2sqrtx dx + 2 int_1^sqrt2 sqrt8 - 4x^2 dx$$\text{Area} = 2 \int_{0}^{1} 2\sqrt{x} dx + 2 \int_{1}^{\sqrt{2}} \sqrt{8 - 4x^2} dx$$ = 4 left[ frac23 x^3/2 right]_0^1 + 4 int_1^sqrt2 sqrt2 - x^2 dx$$= 4 \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} + 4 \int_{1}^{\sqrt{2}} \sqrt{2 - x^2} dx$$ = frac83 + 4 cdot frac12 left[ xsqrt2-x^2 + 2sin^-1left(fracxsqrt2right) right]_1^sqrt2$$= \frac{8}{3} + 4 \cdot \frac{1}{2} \left[ x\sqrt{2-x^2} + 2\sin^{-1}\left(\frac{x}{\sqrt{2}}\right) \right]_{1}^{\sqrt{2}}$$ = frac83 + 2 left[ left(0 + 2 cdot fracpi2right) - left(1 + 2 cdot fracpi4right) right] = frac83 + 2pi - 2 - pi = pi + frac23$$= \frac{8}{3} + 2 \left[ \left(0 + 2 \cdot \frac{\pi}{2}\right) - \left(1 + 2 \cdot \frac{\pi}{4}\right) \right] = \frac{8}{3} + 2\pi - 2 - \pi = \pi + \frac{2}{3}$$ Area = pi + frac23$= \pi + \frac{2}{3}$ sq. units. ### Pattern Recognition Split area integration at intersection x=1$x=1$ between parabola boundary and ellipse boundary. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Integral Calculus

### 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 1. Evaluate alpha$\alpha$: int_0^64 x^1/3 dx = left[ frac34 x^4/3 right]_0^64 = frac34 (256) = 192$$\int_{0}^{64} x^{1/3} dx = \left[ \frac{3}{4} x^{4/3} \right]_0^{64} = \frac{3}{4} (256) = 192$$ int_0^64 [x^1/3] dx = int_0^1 0 dx + int_1^8 1 dx + int_8^27 2 dx + int_27^64 3 dx$$\int_{0}^{64} [x^{1/3}] dx = \int_{0}^{1} 0 dx + \int_{1}^{8} 1 dx + \int_{8}^{27} 2 dx + \int_{27}^{64} 3 dx$$ = 0 + 7(1) + 19(2) + 37(3) = 7 + 38 + 111 = 156$$= 0 + 7(1) + 19(2) + 37(3) = 7 + 38 + 111 = 156$$ Thus, alpha = 192 - 156 = 36$\alpha = 192 - 156 = 36$. ### Step 1: Evaluate Trigonometric Integral Let E = frac1pi int_0^36pi fracsin^2thetasin^6theta + cos^6theta dtheta = frac36pi int_0^pi fracsin^2thetasin^6theta + cos^6theta dtheta$E = \frac{1}{\pi} \int_{0}^{36\pi} \frac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} d\theta = \frac{36}{\pi} \int_{0}^{\pi} \frac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} d\theta$ $ $= frac72pi int_0^pi/2 fracsin^2thetasin^6theta + cos^6theta dtheta$= \frac{72}{\pi} \int_{0}^{\pi/2} \frac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} d\theta$ Let $ Let $J = \int_{0}^{\pi/2} \frac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} d\theta. By King's property, $. By King's property, $J = \int_{0}^{\pi/2} \frac{\cos^2\theta}{\sin^6\theta + \cos^6\theta} d\theta. $. $2J = int_0^pi/2 frac1sin^6theta + cos^6theta dtheta = int_0^infty frac1 + lambda^2lambda^4 - lambda^2 + 1 dlambda = fracpi2 implies J = fracpi4$2J = \int_{0}^{\pi/2} \frac{1}{\sin^6\theta + \cos^6\theta} d\theta = \int_{0}^{\infty} \frac{1 + \lambda^2}{\lambda^4 - \lambda^2 + 1} d\lambda = \frac{\pi}{2} \implies J = \frac{\pi}{4}$ Wait, checking addition: $ Wait, checking addition: $2J = \frac{\pi}{2} \implies J = \frac{\pi}{2} when evaluating $ when evaluating $\int_0^{\infty} \frac{1 + 1/\lambda^2}{\lambda^2 - 1 + 1/\lambda^2} d\lambda = \pi. So $. So $J = \frac{\pi}{2}. ### Step 2: Final Calculation $. ### Step 2: Final Calculation $E = frac72pi times fracpi2 = 36$E = \frac{72}{\pi} \times \frac{\pi}{2} = 36$$ ### Pattern Recognition Break fractional part integral into piecewise constant steps; convert periodic trigonometric integral using symmetry. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Integral Calculus

### Related Formula int e^x [g(x) + g'(x)] dx = e^x g(x) + C$$\int e^x [g(x) + g'(x)] dx = e^x g(x) + C$$ ### Core Logic Rewrite the numerator (2 - x^2)$(2 - x^2)$ as (1 - x^2) + 1$(1 - x^2) + 1$ to split the integral into two recognizable parts matching the e^x [g(x) + g'(x)]$e^x [g(x) + g'(x)]$ form. I = int e^x left( frac(1 - x^2) + 1sqrt1 + x cdot (1 - x)^3/2 right) dx$$I = \int e^x \left( \frac{(1 - x^2) + 1}{\sqrt{1 + x} \cdot (1 - x)^{3/2}} \right) dx$$ = int e^x left( frac1 - x^2sqrt1 + x cdot (1 - x)^3/2 + frac1sqrt1 + x cdot (1 - x)^3/2 right) dx$$= \int e^x \left( \frac{1 - x^2}{\sqrt{1 + x} \cdot (1 - x)^{3/2}} + \frac{1}{\sqrt{1 + x} \cdot (1 - x)^{3/2}} \right) dx$$ ### Step 1: Simplify Terms Simplify the first term: frac1 - x^2sqrt1 + x cdot (1 - x)^3/2 = frac(1 - x)(1 + x)sqrt1 + x cdot (1 - x)sqrt1 - x = fracsqrt1 + xsqrt1 - x = sqrtfrac1 + x1 - x$$\frac{1 - x^2}{\sqrt{1 + x} \cdot (1 - x)^{3/2}} = \frac{(1 - x)(1 + x)}{\sqrt{1 + x} \cdot (1 - x)\sqrt{1 - x}} = \frac{\sqrt{1 + x}}{\sqrt{1 - x}} = \sqrt{\frac{1 + x}{1 - x}}$$ Now verify if the derivative of g(x) = sqrtfrac1 + x1 - x$g(x) = \sqrt{\frac{1 + x}{1 - x}}$ matches the second term. g'(x) = frac12sqrtfrac1+x1-x cdot frac(1-x)(1) - (1+x)(-1)(1-x)^2 = fracsqrt1-x2sqrt1+x cdot frac2(1-x)^2 = frac1sqrt1+x cdot (1-x)^3/2$$g'(x) = \frac{1}{2\sqrt{\frac{1+x}{1-x}}} \cdot \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2} = \frac{\sqrt{1-x}}{2\sqrt{1+x}} \cdot \frac{2}{(1-x)^2} = \frac{1}{\sqrt{1+x} \cdot (1-x)^{3/2}}$$ This perfectly matches the second term. Thus, the integral evaluates to: f(x) = e^x sqrtfrac1 + x1 - x + C$$f(x) = e^x \sqrt{\frac{1 + x}{1 - x}} + C$$ ### Step 2: Apply Boundary Conditions Given f(0) = 0$f(0) = 0$: e^0 sqrtfrac1 + 01 - 0 + C = 0 Rightarrow 1 + C = 0 Rightarrow C = -1$$e^0 \sqrt{\frac{1 + 0}{1 - 0}} + C = 0 \Rightarrow 1 + C = 0 \Rightarrow C = -1$$ So, f(x) = e^x sqrtfrac1 + x1 - x - 1$f(x) = e^x \sqrt{\frac{1 + x}{1 - x}} - 1$. ### Step 3: Evaluate at x = 1/2 fleft(frac12right) = e^1/2 sqrtfrac1 + 1/21 - 1/2 - 1 = e^1/2 sqrtfrac3/21/2 - 1$$f\left(\frac{1}{2}\right) = e^{1/2} \sqrt{\frac{1 + 1/2}{1 - 1/2}} - 1 = e^{1/2} \sqrt{\frac{3/2}{1/2}} - 1$$ = sqrt3e - 1$= \sqrt{3e} - 1$ ### Pattern Recognition Any integral involving e^x$e^x$ multiplied by an algebraic fraction heavily signals the use of int e^x [g(x) + g'(x)] dx$\int e^x [g(x) + g'(x)] dx$. Breaking the numerator into (1-x^2) + 1$(1-x^2) + 1$ is the classic key to unlock this. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Integrals

### Related Formula int_a^b f(x) dx = int_a^b f(a + b - x) dx quad (textKing's Property)$$\int_a^b f(x) dx = \int_a^b f(a + b - x) dx \quad (\text{King's Property})$$ ### Core Logic Let I = int_fracpi24^frac5pi24fracdx1+sqrt[3]tan 2x dots (1)$I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}}\frac{dx}{1+\sqrt[3]{\tan 2x}} \dots (1)$ Apply King's property. Here a + b = fracpi24 + frac5pi24 = frac6pi24 = fracpi4$a + b = \frac{\pi}{24} + \frac{5\pi}{24} = \frac{6\pi}{24} = \frac{\pi}{4}$. Replace x$x$ with fracpi4 - x$\frac{\pi}{4} - x$: I = int_fracpi24^frac5pi24fracdx1+sqrt[3]tan 2left(fracpi4 - xright)$$I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}}\frac{dx}{1+\sqrt[3]{\tan 2\left(\frac{\pi}{4} - x\right)}}$$ ### Step 1: Apply Trigonometric Identity Notice that tan 2left(fracpi4 - xright) = tanleft(fracpi2 - 2xright) = cot 2x$\tan 2\left(\frac{\pi}{4} - x\right) = \tan\left(\frac{\pi}{2} - 2x\right) = \cot 2x$. So, I = int_fracpi24^frac5pi24fracdx1+sqrt[3]cot 2x$I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}}\frac{dx}{1+\sqrt[3]{\cot 2x}}$. Convert cot 2x$\cot 2x$ to frac1tan 2x$\frac{1}{\tan 2x}$: I = int_fracpi24^frac5pi24fracsqrt[3]tan 2xsqrt[3]tan 2x+1 dx dots (2)$$I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}}\frac{\sqrt[3]{\tan 2x}}{\sqrt[3]{\tan 2x}+1} dx \dots (2)$$ ### Step 2: Add the Integrals Add equations (1) and (2): 2I = int_fracpi24^frac5pi24 frac1 + sqrt[3]tan 2x1 + sqrt[3]tan 2x dx = int_fracpi24^frac5pi24 (1) dx$$2I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{1 + \sqrt[3]{\tan 2x}}{1 + \sqrt[3]{\tan 2x}} dx = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} (1) dx$$ 2I = left[ x right]_fracpi24^frac5pi24 = frac5pi24 - fracpi24 = frac4pi24 = fracpi6$$2I = \left[ x \right]_{\frac{\pi}{24}}^{\frac{5\pi}{24}} = \frac{5\pi}{24} - \frac{\pi}{24} = \frac{4\pi}{24} = \frac{\pi}{6}$$ I = frac12 left( fracpi6 right) = fracpi12$$I = \frac{1}{2} \left( \frac{\pi}{6} \right) = \frac{\pi}{12}$$ ### Pattern Recognition Limits of pi/24$\pi/24$ and 5pi/24$5\pi/24$ add up to pi/4$\pi/4$. When dealing with tan(2x)$\tan(2x)$, multiplying by 2$2$ transforms this upper/lower sum to pi/2$\pi/2$, establishing the classic tan/cot$\tan/\cot$ symmetry map triggered by King's Property. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Integrals