Let [cdot]$[\cdot]$ denote the greatest integer function. Then int_-fracpi2^fracpi2left(frac12(3+[x])3+[sin x]+[cos x]right)dx$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{12(3+[x])}{3+[\sin x]+[\cos x]}\right)dx$ is equal to:

### Related Formula int_-a^a f(x) dx = int_0^a [f(x) + f(-x)] dx$$\int_{-a}^{a} f(x) dx = \int_{0}^{a} [f(x) + f(-x)] dx$$ ### Core Logic Let I = int_-pi/6^pi/6fracpi+4x^111-sin(|x|+pi/6)dx$I = \int_{-\pi/6}^{\pi/6}\frac{\pi+4x^{11}}{1-\sin(|x|+\pi/6)}dx$. The denominator 1 - sin(|x| + pi/6)$1 - \sin(|x| + \pi/6)$ is an even function. The numerator can be split into an even part (pi$\pi$) and an odd part (4x^11$4x^{11}$). int_-a^a frac4x^111-sin(|x|+pi/6) dx = 0 quad text(Since integrand is odd)$$\int_{-a}^{a} \frac{4x^{11}}{1-\sin(|x|+\pi/6)} dx = 0 \quad \text{(Since integrand is odd)}$$ ### Step 1: Simplify to Even Integral We are left with the even part: I = int_-pi/6^pi/6 fracpi1 - sin(|x| + pi/6) dx$$I = \int_{-\pi/6}^{\pi/6} \frac{\pi}{1 - \sin(|x| + \pi/6)} dx$$ Using even function property int_-a^a f(x) dx = 2 int_0^a f(x) dx$\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$: I = 2pi int_0^pi/6 frac11 - sin(x + pi/6) dx$$I = 2\pi \int_{0}^{\pi/6} \frac{1}{1 - \sin(x + \pi/6)} dx$$ ### Step 2: Substitution Let t = x + fracpi6 Rightarrow dt = dx$t = x + \frac{\pi}{6} \Rightarrow dt = dx$. Limits: when x = 0 Rightarrow t = pi/6$x = 0 \Rightarrow t = \pi/6$, when x = pi/6 Rightarrow t = pi/3$x = \pi/6 \Rightarrow t = \pi/3$. I = 2pi int_pi/6^pi/3 fracdt1 - sin t$$I = 2\pi \int_{\pi/6}^{\pi/3} \frac{dt}{1 - \sin t}$$ ### Step 3: Solve the Integral Rationalize the denominator: I = 2pi int_pi/6^pi/3 frac1 + sin t(1 - sin t)(1 + sin t) dt$$I = 2\pi \int_{\pi/6}^{\pi/3} \frac{1 + \sin t}{(1 - \sin t)(1 + \sin t)} dt$$ I = 2pi int_pi/6^pi/3 frac1 + sin tcos^2 t dt$$I = 2\pi \int_{\pi/6}^{\pi/3} \frac{1 + \sin t}{\cos^2 t} dt$$ I = 2pi int_pi/6^pi/3 (sec^2 t + sec t tan t) dt$$I = 2\pi \int_{\pi/6}^{\pi/3} (\sec^2 t + \sec t \tan t) dt$$ Integrate directly: I = 2pi left[ tan t + sec t right]_pi/6^pi/3$$I = 2\pi \left[ \tan t + \sec t \right]_{\pi/6}^{\pi/3}$$ Evaluate limits: Upper limit (pi/3$\pi/3$): tan(pi/3) + sec(pi/3) = sqrt3 + 2$\tan(\pi/3) + \sec(\pi/3) = \sqrt{3} + 2$ Lower limit (pi/6$\pi/6$): tan(pi/6) + sec(pi/6) = frac1sqrt3 + frac2sqrt3 = frac3sqrt3 = sqrt3$\tan(\pi/6) + \sec(\pi/6) = \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$ I = 2pi [(sqrt3 + 2) - sqrt3] = 2pi (2) = 4pi$$I = 2\pi [(\sqrt{3} + 2) - \sqrt{3}] = 2\pi (2) = 4\pi$$ ### Pattern Recognition Symmetric limits [-a, a]$[-a, a]$ instantly demand testing for odd/even parity. Any mixed polynomial like c + k x^textodd$c + k x^{\text{odd}}$ over an even denominator guarantees the odd power term strictly vanishes, halving calculation time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Definite Integration Class 11 Maths: Trigonometric Functions

### Related Formula sin A + sin B = 2 sinleft(fracA+B2right) cosleft(fracA-B2right)$$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$ sin 2x = 2 sin x cos x$$\sin 2x = 2 \sin x \cos x$$ cos 2x = 2 cos^2 x - 1$$\cos 2x = 2 \cos^2 x - 1$$ ### Core Logic Let I = 6int_0^pi|sin 3x + sin x + sin 2x| dx$I = 6\int_{0}^{\pi}|\sin 3x + \sin x + \sin 2x| dx$. Apply sum-to-product on sin 3x + sin x$\sin 3x + \sin x$: sin 3x + sin x = 2 sin(2x) cos(x)$\sin 3x + \sin x = 2 \sin(2x) \cos(x)$ So the expression becomes: |2 sin(2x) cos x + sin 2x| = |sin 2x (2 cos x + 1)| = |2 sin x cos x (2 cos x + 1)|$|2 \sin(2x) \cos x + \sin 2x| = |\sin 2x (2 \cos x + 1)| = |2 \sin x \cos x (2 \cos x + 1)|$ Since x in [0, pi]$x \in [0, \pi]$, sin x geq 0$\sin x \geq 0$. We can pull it out of the modulus. I = 12 int_0^pi sin x |2 cos^2 x + cos x| dx$I = 12 \int_{0}^{\pi} \sin x |2 \cos^2 x + \cos x| dx$ ### Step 1: Coordinate Substitution Substitute t = cos x$t = \cos x$, then dt = -sin x dx$dt = -\sin x dx$. Limits: when x = 0$x = 0$, t = 1$t = 1$. When x = pi$x = \pi$, t = -1$t = -1$. I = 12 int_-1^1 |2t^2 + t| dt$$I = 12 \int_{-1}^{1} |2t^2 + t| dt$$ ### Step 2: Resolve Modulus Intervals The roots of 2t^2 + t = 0$2t^2 + t = 0$ are t = 0$t = 0$ and t = -1/2$t = -1/2$. The quadratic 2t^2 + t$2t^2 + t$ is negative in the interval (-1/2, 0)$(-1/2, 0)$ and positive elsewhere. Split the integral: I = 12 left[ int_-1^-1/2 (2t^2 + t) dt - int_-1/2^0 (2t^2 + t) dt + int_0^1 (2t^2 + t) dt right]$$I = 12 \left[ \int_{-1}^{-1/2} (2t^2 + t) dt - \int_{-1/2}^{0} (2t^2 + t) dt + \int_{0}^{1} (2t^2 + t) dt \right]$$

### Step 3: Evaluate Integrals Anti-derivative: F(t) = frac2t^33 + fract^22$F(t) = \frac{2t^3}{3} + \frac{t^2}{2}$. F(1) = 2/3 + 1/2 = 7/6$F(1) = 2/3 + 1/2 = 7/6$ F(0) = 0$F(0) = 0$ F(-1/2) = 2(-1/8)/3 + 1/8 = -1/12 + 1/8 = 1/24$F(-1/2) = 2(-1/8)/3 + 1/8 = -1/12 + 1/8 = 1/24$ F(-1) = -2/3 + 1/2 = -1/6$F(-1) = -2/3 + 1/2 = -1/6$ Evaluate each segment: 1) int_-1^-1/2 = F(-1/2) - F(-1) = 1/24 - (-1/6) = 1/24 + 4/24 = 5/24$\int_{-1}^{-1/2} = F(-1/2) - F(-1) = 1/24 - (-1/6) = 1/24 + 4/24 = 5/24$ 2) -int_-1/2^0 = -(F(0) - F(-1/2)) = -(0 - 1/24) = 1/24$-\int_{-1/2}^{0} = -(F(0) - F(-1/2)) = -(0 - 1/24) = 1/24$ 3) int_0^1 = F(1) - F(0) = 7/6 - 0 = 28/24$\int_{0}^{1} = F(1) - F(0) = 7/6 - 0 = 28/24$ Sum of parts inside bracket: frac524 + frac124 + frac2824 = frac3424 = frac1712$\frac{5}{24} + \frac{1}{24} + \frac{28}{24} = \frac{34}{24} = \frac{17}{12}$ ### Step 4: Final Output I = 12 times frac1712 = 17$$I = 12 \times \frac{17}{12} = 17$$ ### Pattern Recognition Whenever an integral features a cascading sum of sine frequencies like sin(kx)$\sin(kx)$, pair the highest and lowest frequencies first. The resulting common factor often matches the middle term, instantly yielding a clean polynomial substitution under t = cos x$t = \cos x$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Definite Integration Class 11 Maths: Trigonometric Functions

### Related Formula cot^-1(y) = tan^-1left(frac1yright)$$\cot^{-1}(y) = \tan^{-1}\left(\frac{1}{y}\right)$$ tan^-1left(fracx - y1 + xyright) = tan^-1x - tan^-1y$$\tan^{-1}\left(\frac{x - y}{1 + xy}\right) = \tan^{-1}x - \tan^{-1}y$$ textKing's Property: int_0^a f(x)dx = int_0^a f(a-x)dx$$\text{King's Property: } \int_0^a f(x)dx = \int_0^a f(a-x)dx$$ ### Core Logic Let I = int_0^1 cot^-1(1-2x+4x^2) dx$I = \int_{0}^{1} \cot^{-1}(1-2x+4x^2) dx$. Convert cot^-1$\cot^{-1}$ to tan^-1$\tan^{-1}$: cot^-1(1 + 2x(2x-1)) = tan^-1left( frac11 + 2x(2x-1) right)$$\cot^{-1}(1 + 2x(2x-1)) = \tan^{-1}\left( \frac{1}{1 + 2x(2x-1)} \right)$$ Notice that 2x - (2x-1) = 1$2x - (2x-1) = 1$. Thus, the integrand is tan^-1left( frac2x - (2x-1)1 + 2x(2x-1) right)$\tan^{-1}\left( \frac{2x - (2x-1)}{1 + 2x(2x-1)} \right)$. I = int_0^1 left( tan^-1(2x) - tan^-1(2x-1) right) dx$$I = \int_0^1 \left( \tan^{-1}(2x) - \tan^{-1}(2x-1) \right) dx$$ ### Step 1: Apply Definite Integral Properties Applying King's property to the second term int_0^1 tan^-1(2x-1) dx$\int_0^1 \tan^{-1}(2x-1) dx$: x to 1-x implies int_0^1 tan^-1(2(1-x)-1) dx = int_0^1 tan^-1(1-2x) dx = -int_0^1 tan^-1(2x-1) dx$$x \to 1-x \implies \int_0^1 \tan^{-1}(2(1-x)-1) dx = \int_0^1 \tan^{-1}(1-2x) dx = -\int_0^1 \tan^{-1}(2x-1) dx$$ Wait, this implies int_0^1 tan^-1(2x-1) dx = 0$\int_0^1 \tan^{-1}(2x-1) dx = 0$! Thus, I = int_0^1 tan^-1(2x) dx$I = \int_0^1 \tan^{-1}(2x) dx$. ### Step 2: Integration by Parts Solve int_0^1 tan^-1(2x) cdot 1 \, dx$\int_0^1 \tan^{-1}(2x) \cdot 1 \, dx$: I = left[ x tan^-1(2x) right]_0^1 - int_0^1 x frac21+4x^2 dx$$I = \left[ x \tan^{-1}(2x) \right]_0^1 - \int_0^1 x \frac{2}{1+4x^2} dx$$ I = tan^-1(2) - frac14 int_0^1 frac8x1+4x^2 dx$$I = \tan^{-1}(2) - \frac{1}{4} \int_0^1 \frac{8x}{1+4x^2} dx$$ Let 1+4x^2 = t implies 8x dx = dt$1+4x^2 = t \implies 8x dx = dt$. At x=0, t=1$x=0, t=1$; at x=1, t=5$x=1, t=5$. I = tan^-1(2) - frac14 int_1^5 fracdtt = tan^-1(2) - frac14 ln(5)$$I = \tan^{-1}(2) - \frac{1}{4} \int_1^5 \frac{dt}{t} = \tan^{-1}(2) - \frac{1}{4} \ln(5)$$ ### Step 3: Compare and Calculate Result The original integral has a factor of 4$4$: 4I = 4tan^-1(2) - ln(5)$$4I = 4\tan^{-1}(2) - \ln(5)$$ Compare with atan^-1(2) - blog_e(5)$a\tan^{-1}(2) - b\log_e(5)$: a = 4, quad b = 1$$a = 4, \quad b = 1$$ Therefore, 2a + b = 2(4) + 1 = 9$2a + b = 2(4) + 1 = 9$. ### Pattern Recognition Always convert cot^-1$\cot^{-1}$ quadratic inputs into tan^-1fracx-y1+xy$\tan^{-1}\frac{x-y}{1+xy}$ forms. Apply King's property on symmetric limits; often one piece vanishes entirely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Definite Integrals Class 12 Maths: Inverse Trigonometric Functions

### Related Formula textThe greatest integer function [x] text is piecewise constant on intervals [n, n+1).$$\text{The greatest integer function } [x] \text{ is piecewise constant on intervals } [n, n+1).$$ int_a^b f(x) dx text is broken into sub-intervals where f(x) text is constant.$$\int_{a}^{b} f(x) dx \text{ is broken into sub-intervals where } f(x) \text{ is constant.}$$ ### Core Logic The integral bounds are from -pi/2 approx -1.57$-\pi/2 \approx -1.57$ to pi/2 approx 1.57$\pi/2 \approx 1.57$. We must split the integral at every integer point between these bounds. Intervals: 1) [-pi/2, -1) implies [x] = -2$[-\pi/2, -1) \implies [x] = -2$ 2) [-1, 0) implies [x] = -1$[-1, 0) \implies [x] = -1$ 3) [0, 1) implies [x] = 0$[0, 1) \implies [x] = 0$ 4) [1, pi/2) implies [x] = 1$[1, \pi/2) \implies [x] = 1$ ### Step 1: Splitting the Integral I = int_-pi/2^pi/2frac1[x]+4dx$$I = \int_{-\pi/2}^{\pi/2}\frac{1}{[x]+4}dx$$ I = int_-pi/2^-1 frac1-2 + 4 dx + int_-1^0 frac1-1 + 4 dx + int_0^1 frac10 + 4 dx + int_1^pi/2 frac11 + 4 dx$$I = \int_{-\pi/2}^{-1} \frac{1}{-2 + 4} dx + \int_{-1}^{0} \frac{1}{-1 + 4} dx + \int_{0}^{1} \frac{1}{0 + 4} dx + \int_{1}^{\pi/2} \frac{1}{1 + 4} dx$$ ### Step 2: Evaluating Sub-integrals I = int_-pi/2^-1 frac12 dx + int_-1^0 frac13 dx + int_0^1 frac14 dx + int_1^pi/2 frac15 dx$$I = \int_{-\pi/2}^{-1} \frac{1}{2} dx + \int_{-1}^{0} \frac{1}{3} dx + \int_{0}^{1} \frac{1}{4} dx + \int_{1}^{\pi/2} \frac{1}{5} dx$$ Evaluate the limits for each constant integral: I = frac12 left( -1 - left(-fracpi2right) right) + frac13 (0 - (-1)) + frac14 (1 - 0) + frac15 left( fracpi2 - 1 right)$$I = \frac{1}{2} \left( -1 - \left(-\frac{\pi}{2}\right) \right) + \frac{1}{3} (0 - (-1)) + \frac{1}{4} (1 - 0) + \frac{1}{5} \left( \frac{\pi}{2} - 1 \right)$$ I = frac12 left( fracpi2 - 1 right) + frac13 + frac14 + frac15 left( fracpi2 - 1 right)$$I = \frac{1}{2} \left( \frac{\pi}{2} - 1 \right) + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \left( \frac{\pi}{2} - 1 \right)$$ ### Step 3: Simplifying the Expression Group the left( fracpi2 - 1 right)$\left( \frac{\pi}{2} - 1 \right)$ terms: I = left(frac12 + frac15right) left( fracpi2 - 1 right) + frac13 + frac14$$I = \left(\frac{1}{2} + \frac{1}{5}\right) \left( \frac{\pi}{2} - 1 \right) + \frac{1}{3} + \frac{1}{4}$$ I = frac710 left( fracpi2 - 1 right) + frac712$$I = \frac{7}{10} \left( \frac{\pi}{2} - 1 \right) + \frac{7}{12}$$ I = frac7pi20 - frac710 + frac712$$I = \frac{7\pi}{20} - \frac{7}{10} + \frac{7}{12}$$ Find a common denominator for the constants (LCD is 60): -frac710 + frac712 = -frac4260 + frac3560 = -frac760$$-\frac{7}{10} + \frac{7}{12} = -\frac{42}{60} + \frac{35}{60} = -\frac{7}{60}$$ Rewrite frac7pi20$\frac{7\pi}{20}$ with denominator 60: frac7pi20 = frac21pi60$$\frac{7\pi}{20} = \frac{21\pi}{60}$$ So, I = frac21pi60 - frac760 = frac7(3pi - 1)60 = frac760(3pi - 1)$$I = \frac{21\pi}{60} - \frac{7}{60} = \frac{7(3\pi - 1)}{60} = \frac{7}{60}(3\pi - 1)$$ ### Pattern Recognition Integration of step functions always transforms into a simple sum of rectangle areas (c_i times Delta x_i)$(c_i \times \Delta x_i)$. Immediately break the bounds at integers, turning a calculus problem into elementary arithmetic. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Definite Integration

### Related Formula Integration by parts formula: int u \, dv = uv - int v \, du$\int u \, dv = uv - \int v \, du$. ### Core Logic Let I(x) = int_0^x t^2 sin(x - t) dt$I(x) = \int_{0}^{x} t^2 \sin(x - t) dt$. Use integration by parts. Let u = t^2$u = t^2$ and dv = sin(x - t) dt implies v = cos(x - t)$dv = \sin(x - t) dt \implies v = \cos(x - t)$. I(x) = [t^2 cos(x - t)]_0^x - int_0^x 2t cos(x - t) dt$$I(x) = [t^2 \cos(x - t)]_{0}^{x} - \int_{0}^{x} 2t \cos(x - t) dt$$ I(x) = x^2 cos(0) - 0 - int_0^x 2t cos(x - t) dt$$I(x) = x^2 \cos(0) - 0 - \int_{0}^{x} 2t \cos(x - t) dt$$ I(x) = x^2 - int_0^x 2t cos(x - t) dt$$I(x) = x^2 - \int_{0}^{x} 2t \cos(x - t) dt$$ ### Step 1: Second Integration by Parts Now integrate int_0^x 2t cos(x - t) dt$\int_{0}^{x} 2t \cos(x - t) dt$ by parts again: Let u = 2t$u = 2t$ and dv = cos(x - t) dt implies v = -sin(x - t)$dv = \cos(x - t) dt \implies v = -\sin(x - t)$. = [2t(-sin(x - t))]_0^x - int_0^x 2(-sin(x - t)) dt$$= [2t(-\sin(x - t))]_{0}^{x} - \int_{0}^{x} 2(-\sin(x - t)) dt$$ = (-2xsin(0) - 0) + int_0^x 2sin(x - t) dt$$= (-2x\sin(0) - 0) + \int_{0}^{x} 2\sin(x - t) dt$$ = 0 + [2cos(x - t)]_0^x = 2cos(0) - 2cos(x) = 2 - 2cos x$$= 0 + [2\cos(x - t)]_{0}^{x} = 2\cos(0) - 2\cos(x) = 2 - 2\cos x$$ ### Step 2: Equating and Solving Substitute this back into the first equation: I(x) = x^2 - (2 - 2cos x) = x^2 + 2cos x - 2$$I(x) = x^2 - (2 - 2\cos x) = x^2 + 2\cos x - 2$$ We are given I(x) = x^2$I(x) = x^2$. Therefore: x^2 + 2cos x - 2 = x^2$$x^2 + 2\cos x - 2 = x^2$$ 2cos x = 2 implies cos x = 1$$2\cos x = 2 \implies \cos x = 1$$ The solutions for cos x = 1$\cos x = 1$ are x = 2npi$x = 2n\pi$, where n$n$ is an integer. We need the roots in the interval [0, 100]$[0, 100]$. 0 leq 2npi leq 100 implies 0 leq n leq frac1002pi approx frac1006.28 approx 15.92$0 \leq 2n\pi \leq 100 \implies 0 \leq n \leq \frac{100}{2\pi} \approx \frac{100}{6.28} \approx 15.92$. Since n$n$ must be an integer, n$n$ takes values 0, 1, 2, dots, 15$0, 1, 2, \dots, 15$. The total number of solutions is 16$16$. ### Pattern Recognition Convoluted-looking definite integrals containing a shift (x-t)$(x-t)$ often rapidly unpack through successive integration by parts where the polynomial variable diminishes until exhaustion. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Definite Integration Class 11 Maths: Trigonometric Equations