The area of the region A = \(x,y) : 4x^2 + y^2 le 8 text and y^2 le 4x\$A = \{(x,y) : 4x^2 + y^2 \le 8 \text{ and } y^2 \le 4x\}$ is:

### 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 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 textKing's Property: int_a^b f(x) dx = int_a^b f(a+b-x) dx$$\text{King's Property: } \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$$ ### Core Logic First, we find the coordinates of intersection in the second quadrant to determine the integration limits a$a$ and b$b$. ### Step 1: Find points of intersection Substitute y = fracx^22$y = \frac{x^2}{2}$ into the line equation y - 2x - 6 = 0$y - 2x - 6 = 0$: fracx^22 - 2x - 6 = 0 implies x^2 - 4x - 12 = 0$$\frac{x^2}{2} - 2x - 6 = 0 \implies x^2 - 4x - 12 = 0$$ (x - 6)(x + 2) = 0 implies x = 6 quad textor quad x = -2$$(x - 6)(x + 2) = 0 \implies x = 6 \quad \text{or} \quad x = -2$$ Since the point (a, b)$(a, b)$ lies in the second quadrant, x$x$ must be negative: a = -2$a = -2$ b = 2(a) + 6 = 2(-2) + 6 = 2$$b = 2(a) + 6 = 2(-2) + 6 = 2$$ Thus, the integration limits are a = -2$a = -2$ and b = 2$b = 2$. ### Step 2: Solve the Integral using King's Property The integral is: I = int_-2^2 frac9x^21 + 5^x dx quad text--- (1)$$I = \int_{-2}^{2} \frac{9x^2}{1 + 5^x} dx \quad \text{--- (1)}$$ Apply King's property, substituting x to -x$x \to -x$ (since -2 + 2 - x = -x$-2 + 2 - x = -x$): I = int_-2^2 frac9(-x)^21 + 5^-x dx = int_-2^2 frac9x^2 cdot 5^x1 + 5^x dx quad text--- (2)$$I = \int_{-2}^{2} \frac{9(-x)^2}{1 + 5^{-x}} dx = \int_{-2}^{2} \frac{9x^2 \cdot 5^x}{1 + 5^x} dx \quad \text{--- (2)}$$ Adding equations (1) and (2): 2I = int_-2^2 9x^2 left( frac1 + 5^x1 + 5^x right) dx = int_-2^2 9x^2 dx$$2I = \int_{-2}^{2} 9x^2 \left( \frac{1 + 5^x}{1 + 5^x} \right) dx = \int_{-2}^{2} 9x^2 dx$$ Since 9x^2$9x^2$ is an even function: 2I = 2 int_0^2 9x^2 dx implies I = left[ 3x^3 right]_0^2 = 3(8) - 0 = 24$$2I = 2 \int_{0}^{2} 9x^2 dx \implies I = \left[ 3x^3 \right]_{0}^{2} = 3(8) - 0 = 24$$ ### Pattern Recognition Whenever you see an exponential denominator like 1 + c^x$1 + c^x$ inside a symmetric interval definite integral [-a, a]$[-a, a]$, applying King's property will almost always cancel the exponential factor cleanly when the remaining numerator is even. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus

### Related Formula int sqrta^2 + x^2 dx = fracx2 sqrta^2 + x^2 + fraca^22 lnleft| x + sqrta^2 + x^2 right|$$\int \sqrt{a^2 + x^2} dx = \frac{x}{2} \sqrt{a^2 + x^2} + \frac{a^2}{2} \ln\left| x + \sqrt{a^2 + x^2} \right|$$ ### Core Logic We first rationalise the denominator to split the integral into two standard integration terms. ### Step 1: Rationalise the integrand Multiply the numerator and denominator by sqrt3+x^2 - sqrt1+x^2$\sqrt{3+x^2} - \sqrt{1+x^2}$: frac1sqrt3 + x^2 + sqrt1 + x^2 = fracsqrt3 + x^2 - sqrt1 + x^2(3+x^2) - (1+x^2) = fracsqrt3 + x^2 - sqrt1 + x^22$$\frac{1}{\sqrt{3 + x^2} + \sqrt{1 + x^2}} = \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{(3+x^2) - (1+x^2)} = \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{2}$$ Thus, the integral expression simplifies to: I = 4 int_0^1 left( fracsqrt3 + x^2 - sqrt1 + x^22 right) dx - 3log_eleft(sqrt3right)$$I = 4 \int_{0}^{1} \left( \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{2} \right) dx - 3\log_{e}\left(\sqrt{3}\right)$$ I = 2 int_0^1 sqrt3 + x^2 dx - 2 int_0^1 sqrt1 + x^2 dx - frac32 log_e 3$$I = 2 \int_{0}^{1} \sqrt{3 + x^2} dx - 2 \int_{0}^{1} \sqrt{1 + x^2} dx - \frac{3}{2} \log_{e} 3$$ ### Step 2: Evaluate the integrals For the first integral: 2 int_0^1 sqrt3 + x^2 dx = 2 left[ fracx2 sqrt3 + x^2 + frac32 lnleft| x + sqrt3 + x^2 right| right]_0^1$$2 \int_{0}^{1} \sqrt{3 + x^2} dx = 2 \left[ \frac{x}{2} \sqrt{3 + x^2} + \frac{3}{2} \ln\left| x + \sqrt{3 + x^2} \right| \right]_{0}^{1}$$ = left[ x sqrt3 + x^2 + 3 lnleft| x + sqrt3 + x^2 right| right]_0^1$$= \left[ x \sqrt{3 + x^2} + 3 \ln\left| x + \sqrt{3 + x^2} \right| \right]_{0}^{1}$$ = left( sqrt4 + 3 ln(1 + sqrt4) right) - left( 0 + 3 lnsqrt3 right)$$= \left( \sqrt{4} + 3 \ln(1 + \sqrt{4}) \right) - \left( 0 + 3 \ln\sqrt{3} \right)$$ = 2 + 3 ln 3 - frac32 ln 3 = 2 + frac32 ln 3$$= 2 + 3 \ln 3 - \frac{3}{2} \ln 3 = 2 + \frac{3}{2} \ln 3$$ For the second integral: -2 int_0^1 sqrt1 + x^2 dx = -2 left[ fracx2 sqrt1 + x^2 + frac12 lnleft| x + sqrt1 + x^2 right| right]_0^1$$-2 \int_{0}^{1} \sqrt{1 + x^2} dx = -2 \left[ \frac{x}{2} \sqrt{1 + x^2} + \frac{1}{2} \ln\left| x + \sqrt{1 + x^2} \right| \right]_{0}^{1}$$ = - left[ x sqrt1 + x^2 + lnleft| x + sqrt1 + x^2 right| right]_0^1$$= - \left[ x \sqrt{1 + x^2} + \ln\left| x + \sqrt{1 + x^2} \right| \right]_{0}^{1}$$ = - left( sqrt2 + ln(1 + sqrt2) right)$$= - \left( \sqrt{2} + \ln(1 + \sqrt{2}) \right)$$ ### Step 3: Sum the terms Now compile all terms: I = left( 2 + frac32 ln 3 right) - sqrt2 - ln(1 + sqrt2) - frac32 ln 3$$I = \left( 2 + \frac{3}{2} \ln 3 \right) - \sqrt{2} - \ln(1 + \sqrt{2}) - \frac{3}{2} \ln 3$$ I = 2 - sqrt2 - ln(1 + sqrt2)$$I = 2 - \sqrt{2} - \ln(1 + \sqrt{2})$$ ### Pattern Recognition Integration of roots of quadratics: Always look for algebraic rationalisation when dealing with sum of root denominators. It directly reduces complex fractions into standard integrable functions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus