The value of int_-fracpi2^fracpi2left(frac1[x]+4right)dx$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{1}{[x]+4}\right)dx$, where [bullet]$[\bullet]$ denotes the greatest integer function, is

### 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 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 = int_0^pi frac(x + 3)sin x1 + 3cos^2x dx quad dots (1)$$I = \int_0^\pi \frac{(x + 3)\sin x}{1 + 3\cos^2x} dx \quad \dots (1)$$ Applying King's property (x to pi - x$x \to \pi - x$): I = int_0^pi frac(pi - x + 3)sin(pi - x)1 + 3cos^2(pi - x) dx$$I = \int_0^\pi \frac{(\pi - x + 3)\sin(\pi - x)}{1 + 3\cos^2(\pi - x)} dx$$ I = int_0^pi frac(pi - x + 3)sin x1 + 3cos^2x dx quad dots (2)$$I = \int_0^\pi \frac{(\pi - x + 3)\sin x}{1 + 3\cos^2x} dx \quad \dots (2)$$ ### Step 1: Eliminate the x Variable Adding equations (1) and (2): 2I = int_0^pi frac[(x + 3) + (pi - x + 3)]sin x1 + 3cos^2x dx$$2I = \int_0^\pi \frac{[(x + 3) + (\pi - x + 3)]\sin x}{1 + 3\cos^2x} dx$$ 2I = (pi + 6)int_0^pi fracsin x1 + 3cos^2x dx$$2I = (\pi + 6)\int_0^\pi \frac{\sin x}{1 + 3\cos^2x} dx$$ Using the symmetric property int_0^2a f(x)dx = 2int_0^a f(x)dx$\int_0^{2a} f(x)dx = 2\int_0^a f(x)dx$ if f(2a-x)=f(x)$f(2a-x)=f(x)$: 2I = 2(pi + 6)int_0^pi/2 fracsin x1 + 3cos^2x dx$$2I = 2(\pi + 6)\int_0^{\pi/2} \frac{\sin x}{1 + 3\cos^2x} dx$$ I = (pi + 6)int_0^pi/2 fracsin x1 + 3cos^2x dx$$I = (\pi + 6)\int_0^{\pi/2} \frac{\sin x}{1 + 3\cos^2x} dx$$ ### Step 2: Solve Using Substitution Let t = sqrt3cos x$t = \sqrt{3}\cos x$. Then dt = -sqrt3sin x dx implies sin x dx = -fracdtsqrt3$dt = -\sqrt{3}\sin x dx \implies \sin x dx = -\frac{dt}{\sqrt{3}}$. Change in integration boundaries: - When x = 0 implies t = sqrt3$x = 0 \implies t = \sqrt{3}$ - When x = pi/2 implies t = 0$x = \pi/2 \implies t = 0$ Substituting into the integral: I = (pi + 6) int_sqrt3^0 frac-dt/sqrt31 + t^2 = fracpi + 6sqrt3 int_0^sqrt3 fracdt1 + t^2$$I = (\pi + 6) \int_{\sqrt{3}}^0 \frac{-dt/\sqrt{3}}{1 + t^2} = \frac{\pi + 6}{\sqrt{3}} \int_0^{\sqrt{3}} \frac{dt}{1 + t^2}$$ I = fracpi + 6sqrt3 left[ tan^-1t right]_0^sqrt3 = fracpi + 6sqrt3 left( tan^-1sqrt3 - 0 right)$$I = \frac{\pi + 6}{\sqrt{3}} \left[ \tan^{-1}t \right]_0^{\sqrt{3}} = \frac{\pi + 6}{\sqrt{3}} \left( \tan^{-1}\sqrt{3} - 0 \right)$$ I = fracpi + 6sqrt3 cdot fracpi3 = fracpi3sqrt3(pi + 6)$$I = \frac{\pi + 6}{\sqrt{3}} \cdot \frac{\pi}{3} = \frac{\pi}{3\sqrt{3}}(\pi + 6)$$ ### Pattern Recognition Whenever you encounter a linear x$x$ factor multiplying trigonometric components in a definite integral with symmetric limits like 0$0$ to pi$\pi$, executing King's property first is almost guaranteed to cleanly wipe out that variable element. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals