Solution & Explanation
### 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
More Definite Integrals Previous-Year Questions — Page 2
Q71
jee_main_2025_29_jan_evening
Integration of Modulus and Greatest Integer Functions
If 24int_0^fracpi4left(sin left|4x - fracpi12right| + [2sin x]right)mathrmdx = 2pi +alpha$24\int_{0}^{\frac{pi}{4}}\left(\sin \left|4x - \frac{pi}{12}\right| + [2\sin x]\right)\mathrm{d}x = 2\pi +\alpha$, where [cdot ]$[\cdot ]$ denotes the greatest integer function, then alpha$\alpha$ is equal to
Numerical Answer. Answer: 12 to 12
Solution
### 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
Q63
jee_main_2025_28_jan_morning
Properties of Definite Integrals (King's Property)
If int_-fracpi2^fracpi2frac96x^2cos^2x(1 + e^x) dx = pi (alpha pi^2 +beta),alpha ,beta in mathbbZ,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{96x^2\cos^2x}{(1 + e^x)} dx = \pi (\alpha \pi^2 +\beta),\alpha ,\beta \in \mathbb{Z},$ then (alpha + beta)^2$(\alpha + \beta)^2$ equals:
(1) 144
(2) 196
(3) 100
(4) 64
- A. 144
- B. 196
- C. 100
- D. 64
Solution
### 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
Q68
jee_main_2025_04_april_evening
Properties of Definite Integrals
Let f(x) + 2fleft(frac1xright) = x^2 + 5$f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5$ and 2 mathrm g (mathrm x) - 3 mathrm g left(frac 12right) = mathrm x, mathrm x > 0$2 \mathrm {g} (\mathrm {x}) - 3 \mathrm {g} \left(\frac {1}{2}\right) = \mathrm {x}, \mathrm {x} > 0$. If alpha = int_ 1 ^ 2 f (x) d x$\alpha = \int_ {1} ^ {2} f (x) d x$, and beta = int_ 1 ^ 2 g (x) d x$\beta = \int_ {1} ^ {2} g (x) d x$, then the value of 9alpha + beta$9\alpha + \beta$ is:
- A. 1$1$
- B. 0$0$
- C. 10$10$
- D. 11$11$
Solution
### 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
Q66
jee_main_2025_04_april_morning
Properties of Definite Integrals
The value of int_-1^1fracleft(1 + sqrt|x| - xright)e^x + left(sqrt|x| - xright)e^-xe^x + e^-x \, mathrmdx$\int_{-1}^{1}\frac{\left(1 + \sqrt{|x| - x}\right)e^x + \left(\sqrt{|x| - x}\right)e^{-x}}{e^x + e^{-x}} \, \mathrm{d}x$ is equal to
- A. 3 - frac2sqrt23$3 - \frac{2\sqrt{2}}{3}$
- B. 2 + frac2sqrt23$2 + \frac{2\sqrt{2}}{3}$
- C. 1 - frac2sqrt23$1 - \frac{2\sqrt{2}}{3}$
- D. 1 + frac2sqrt23$1 + \frac{2\sqrt{2}}{3}$
Solution
### 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
Q52
jee_main_2025_24_jan_morning
Properties of Definite Integrals
If I(m,n) = int_0^1 x^m-1 (1-x)^n-1 dx$I(m,n) = \int_{0}^{1} x^{m-1} (1-x)^{n-1} dx$ where m, n > 0$m, n > 0$, then I(9,14) + I(10,13)$I(9,14) + I(10,13)$ is :
- A. I(9, 1)$I(9, 1)$
- B. I(19, 27)$I(19, 27)$
- C. I(1, 13)$I(1, 13)$
- D. I(9, 13)$I(9, 13)$
Solution
### 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
