Fundamental definite integration Questions in English

(B) To evaluate $\int_{0}^{2} [x^{2}] dx$,we split the interval $[0, 2]$ based on the points where $x^{2}$ is an integer: $x^{2} = 1, 2, 3, 4$.
This gives the points $x = 1, \sqrt{2}, \sqrt{3}, 2$.
Thus,the integral becomes:
$\int_{0}^{1} [x^{2}] dx + \int_{1}^{\sqrt{2}} [x^{2}] dx + \int_{\sqrt{2}}^{\sqrt{3}} [x^{2}] dx + \int_{\sqrt{3}}^{2} [x^{2}] dx$
$= \int_{0}^{1} 0 dx + \int_{1}^{\sqrt{2}} 1 dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 dx + \int_{\sqrt{3}}^{2} 3 dx$
$= 0 + [x]_{1}^{\sqrt{2}} + 2[x]_{\sqrt{2}}^{\sqrt{3}} + 3[x]_{\sqrt{3}}^{2}$
$= (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3})$
$= \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3}$
$= 5 - \sqrt{2} - \sqrt{3}$.

(D) We have $I_{1} = \int_{0}^{n} [x] dx = \sum_{k=0}^{n-1} \int_{k}^{k+1} k dx = \sum_{k=0}^{n-1} k(k+1-k) = \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2}$.
Now,$I_{2} = \int_{0}^{n} \{x\} dx$. Since $\{x\} = x - [x]$,we have $I_{2} = \int_{0}^{n} x dx - \int_{0}^{n} [x] dx$.
$I_{2} = \left[ \frac{x^2}{2} \right]_{0}^{n} - I_{1} = \frac{n^2}{2} - \frac{n(n-1)}{2} = \frac{n^2 - n^2 + n}{2} = \frac{n}{2}$.
Therefore,$\frac{I_{1}}{I_{2}} = \frac{\frac{n(n-1)}{2}}{\frac{n}{2}} = n-1$.

(B) Let $I = \int_{0}^{2} x^{2}[x] d x$.
Since $[x]$ is the greatest integer function,we split the integral at the integer points in the interval $[0, 2]$.
$I = \int_{0}^{1} x^{2}[x] d x + \int_{1}^{2} x^{2}[x] d x$.
For $0 \le x < 1$,$[x] = 0$.
For $1 \le x < 2$,$[x] = 1$.
Substituting these values,we get:
$I = \int_{0}^{1} x^{2}(0) d x + \int_{1}^{2} x^{2}(1) d x$.
$I = 0 + \int_{1}^{2} x^{2} d x$.
$I = \left[ \frac{x^{3}}{3} \right]_{1}^{2}$.
$I = \frac{2^{3}}{3} - \frac{1^{3}}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$.

(C) Let $I = \int_{0}^{\sqrt{3}} \{x^2\} dx$.
Since $\{x^2\} = x^2 - [x^2]$,we split the integral based on the values of $[x^2]$:
For $0 \le x < 1$,$[x^2] = 0$.
For $1 \le x < \sqrt{2}$,$[x^2] = 1$.
For $\sqrt{2} \le x < \sqrt{3}$,$[x^2] = 2$.
Thus,$I = \int_{0}^{1} x^2 dx + \int_{1}^{\sqrt{2}} (x^2 - 1) dx + \int_{\sqrt{2}}^{\sqrt{3}} (x^2 - 2) dx$.
$I = \left[ \frac{x^3}{3} \right]_{0}^{1} + \left[ \frac{x^3}{3} - x \right]_{1}^{\sqrt{2}} + \left[ \frac{x^3}{3} - 2x \right]_{\sqrt{2}}^{\sqrt{3}}$.
$I = \left( \frac{1}{3} - 0 \right) + \left( (\frac{2\sqrt{2}}{3} - \sqrt{2}) - (\frac{1}{3} - 1) \right) + \left( (\frac{3\sqrt{3}}{3} - 2\sqrt{3}) - (\frac{2\sqrt{2}}{3} - 2\sqrt{2}) \right)$.
$I = \frac{1}{3} + (-\frac{\sqrt{2}}{3} + \frac{2}{3}) + (-\sqrt{3} + \frac{4\sqrt{2}}{3})$.
$I = \frac{1}{3} + \frac{2}{3} - \sqrt{3} + \frac{3\sqrt{2}}{3} = 1 - \sqrt{3} + \sqrt{2} = \sqrt{2} - \sqrt{3} + 1$.

(A) Given,$f(x) = \begin{cases} 2x^2 + 1, & x \leq 1 \\ 4x^3 - 1, & x > 1 \end{cases}$.
We need to evaluate $\int_{0}^{2} f(x) dx$.
Since the function definition changes at $x = 1$,we split the integral:
$\int_{0}^{2} f(x) dx = \int_{0}^{1} (2x^2 + 1) dx + \int_{1}^{2} (4x^3 - 1) dx$.
Evaluating the first part:
$\int_{0}^{1} (2x^2 + 1) dx = \left[ \frac{2x^3}{3} + x \right]_{0}^{1} = \left( \frac{2(1)^3}{3} + 1 \right) - (0) = \frac{2}{3} + 1 = \frac{5}{3}$.
Evaluating the second part:
$\int_{1}^{2} (4x^3 - 1) dx = \left[ x^4 - x \right]_{1}^{2} = (2^4 - 2) - (1^4 - 1) = (16 - 2) - (0) = 14$.
Adding both parts:
$\int_{0}^{2} f(x) dx = \frac{5}{3} + 14 = \frac{5 + 42}{3} = \frac{47}{3}$.

(C) Given,$f(x) = \max \{x+|x|, x-[x]\}$.
For $x \geq 0$,$x+|x| = 2x$ and $x-[x] = \{x\} \in [0, 1)$. Since $2x \geq \{x\}$ for $x \geq 0$,$f(x) = 2x$.
For $x < 0$,$x+|x| = x-x = 0$ and $x-[x] = \{x\} \in [0, 1)$. Since $0 \leq \{x\} < 1$,$f(x) = x-[x] = \{x\}$.
Thus,$\int_{-3}^{3} f(x) dx = \int_{-3}^{0} (x-[x]) dx + \int_{0}^{3} 2x dx$.
Since $x-[x]$ is periodic with period $1$,$\int_{-3}^{0} (x-[x]) dx = 3 \int_{0}^{1} x dx = 3 \left[ \frac{x^2}{2} \right]_0^1 = \frac{3}{2}$.
And $\int_{0}^{3} 2x dx = [x^2]_0^3 = 9$.
Therefore,$\int_{-3}^{3} f(x) dx = \frac{3}{2} + 9 = \frac{21}{2}$.

(B) Let $g(x) = \lim _{c \rightarrow 0} \int_c^x \frac{b t \cos 4 t - a \sin 4 t}{t^2} d t = \frac{a \sin 4 x}{x} - 1$.
Applying the limit as $x \rightarrow 0$,we have $g(0) = \lim _{x \rightarrow 0} (\frac{a \sin 4 x}{x} - 1) = 4a - 1$.
Since the integral from $c$ to $c$ is $0$,we must have $g(0) = 0$,so $4a - 1 = 0$,which gives $a = 1/4$.
Now,differentiate both sides with respect to $x$ using the Fundamental Theorem of Calculus:
$g'(x) = \frac{b x \cos 4 x - a \sin 4 x}{x^2}$.
Also,differentiating the right side:
$g'(x) = \frac{d}{dx} (\frac{a \sin 4 x}{x} - 1) = \frac{4ax \cos 4 x - a \sin 4 x}{x^2}$.
Comparing the two expressions for $g'(x)$,we get $b = 4a$.
Since $a = 1/4$,we have $b = 4(1/4) = 1$.
Thus,$a = 1/4$ and $b = 1$.

(D) For $0 < x < 1$,we have $x^2 < x$ and $0 \le \cos^2 x \le 1$.
Comparing the integrands:
$e^{-x} \cos^2 x < e^{-x^2} \cos^2 x < e^{-x^2} < e^{-x^2/2}$.
Since the integrand of $I_4$ is the largest for all $x \in (0, 1)$,the integral $I_4$ will be the greatest.
Therefore,$I = I_4$.

(C) For $x \in [10, 19]$,we have $|\sin x| \leq 1$ and $1+x^{6} > 10^{6}$.
Since $x \geq 10$,$1+x^{6} > 10^{6}$,which implies $\frac{1}{1+x^{6}} < 10^{-6}$.
Therefore,$|I| = \left| \int_{10}^{19} \frac{\sin x}{1+x^{6}} dx \right| \leq \int_{10}^{19} \frac{|\sin x|}{1+x^{6}} dx$.
Since $|\sin x| \leq 1$ and $\frac{1}{1+x^{6}} < 10^{-6}$,we have $|I| < \int_{10}^{19} 10^{-6} dx$.
$|I| < 10^{-6} \times (19 - 10) = 9 \times 10^{-6}$.
However,checking the options provided and the magnitude,$9 \times 10^{-6} < 10^{-5}$. Thus,$|I| < 10^{-5}$ is the correct bound.

(D) We know that for $x \in [0, 1]$,$0 \leq x^2 \leq 1$.
Since $e^x$ is an increasing function,we have $e^0 \leq e^{x^2} \leq e^1$,which implies $1 \leq e^{x^2} \leq e$.
Integrating the inequality over the interval $[0, 1]$:
$\int_{0}^{1} 1 \, dx \leq \int_{0}^{1} e^{x^2} \, dx \leq \int_{0}^{1} e \, dx$.
Calculating the integrals:
$[x]_{0}^{1} \leq \int_{0}^{1} e^{x^2} \, dx \leq [ex]_{0}^{1}$.
$1 \leq \int_{0}^{1} e^{x^2} \, dx \leq e$.
Thus,the value of the integral lies in the closed interval $[1, e]$.

(C) Let $I = \int_{1}^{5} [|x-3| + |1-x|] dx$.
Since $x \in [1, 5]$,$|1-x| = x-1$.
Thus,$I = \int_{1}^{5} |x-3| dx + \int_{1}^{5} (x-1) dx$.
For the first part,$\int_{1}^{5} |x-3| dx = \int_{1}^{3} -(x-3) dx + \int_{3}^{5} (x-3) dx$.
$= \int_{1}^{3} (3-x) dx + \int_{3}^{5} (x-3) dx = [3x - \frac{x^2}{2}]_{1}^{3} + [\frac{x^2}{2} - 3x]_{3}^{5}$.
$= (9 - 4.5) - (3 - 0.5) + (12.5 - 15) - (4.5 - 9) = 4.5 - 2.5 - 2.5 + 4.5 = 4$.
For the second part,$\int_{1}^{5} (x-1) dx = [\frac{x^2}{2} - x]_{1}^{5} = (12.5 - 5) - (0.5 - 1) = 7.5 - (-0.5) = 8$.
Therefore,$I = 4 + 8 = 12$.

(C) The given expression is a Riemann sum for the definite integral of the function $f(x)$ over the interval $[0, 1]$.
By the definition of the definite integral as the limit of a sum:
$\lim_{n \rightarrow \infty} \sum_{j=0}^n \frac{1}{n} f\left(\frac{j}{n}\right) = \int_{0}^{1} f(x) dx$
Here,we substitute $\frac{j}{n} = x$ and $\frac{1}{n} = dx$.
The lower limit is $\lim_{n \rightarrow \infty} \frac{0}{n} = 0$.
The upper limit is $\lim_{n \rightarrow \infty} \frac{n}{n} = 1$.
Therefore,the limit is equal to $\int_{0}^{1} f(x) dx$.

(B) Given $f(x) = \max \{x + |x|, x - [x]\}$.
We know that $x - [x] = \{x\}$,where $\{x\}$ is the fractional part of $x$.
For $x \in [-3, 0)$,$x + |x| = x - x = 0$. Since $\{x\} \ge 0$,$f(x) = \max \{0, \{x\}\} = \{x\}$.
For $x \in [0, 3]$,$x + |x| = x + x = 2x$. Since $2x \ge \{x\}$ for $x \ge 0$,$f(x) = 2x$.
Now,calculate the integral:
$\int_{-3}^3 f(x) \, dx = \int_{-3}^0 \{x\} \, dx + \int_0^3 2x \, dx$
$\int_{-3}^0 \{x\} \, dx = \int_{-3}^0 (x - [x]) \, dx$. Since the integral of the fractional part over an interval of length $n$ is $\frac{n}{2}$,$\int_{-3}^0 \{x\} \, dx = 3 \times \frac{1}{2} = \frac{3}{2}$.
$\int_0^3 2x \, dx = [x^2]_0^3 = 9 - 0 = 9$.
Thus,$\int_{-3}^3 f(x) \, dx = \frac{3}{2} + 9 = \frac{21}{2}$.

(D) Let $I_1 = \int_0^n [x] dx$ and $I_2 = \int_0^n \{x\} dx$.
$I_1 = \int_0^1 0 dx + \int_1^2 1 dx + \int_2^3 2 dx + \dots + \int_{n-1}^n (n-1) dx$.
$I_1 = 0 + 1 + 2 + \dots + (n-1) = \frac{(n-1)n}{2}$.
Since the function $\{x\}$ is periodic with period $1$,we have $I_2 = n \int_0^1 \{x\} dx = n \int_0^1 x dx$.
$I_2 = n \left[ \frac{x^2}{2} \right]_0^1 = n \left( \frac{1}{2} - 0 \right) = \frac{n}{2}$.
Therefore,the expression is $\frac{I_1}{I_2} = \frac{\frac{n(n-1)}{2}}{\frac{n}{2}} = n-1$.

(C) To evaluate $\int_{0}^{5} \max \{x^{2}, 6x-8\} dx$,we first find the intersection points of $y = x^{2}$ and $y = 6x-8$.
Setting $x^{2} = 6x-8$,we get $x^{2}-6x+8 = 0$,which factors as $(x-2)(x-4) = 0$. Thus,the curves intersect at $x = 2$ and $x = 4$.
For $x \in [0, 2]$,$x^{2} \ge 6x-8$.
For $x \in [2, 4]$,$6x-8 \ge x^{2}$.
For $x \in [4, 5]$,$x^{2} \ge 6x-8$.
Therefore,the integral is split as:
$\int_{0}^{2} x^{2} dx + \int_{2}^{4} (6x-8) dx + \int_{4}^{5} x^{2} dx$
$= \left[ \frac{x^{3}}{3} \right]_{0}^{2} + \left[ 3x^{2}-8x \right]_{2}^{4} + \left[ \frac{x^{3}}{3} \right]_{4}^{5}$
$= (\frac{8}{3} - 0) + ((48-32) - (12-16)) + (\frac{125}{3} - \frac{64}{3})$
$= \frac{8}{3} + (16 - (-4)) + \frac{61}{3}$
$= \frac{8}{3} + 20 + \frac{61}{3} = \frac{69}{3} + 20 = 23 + 20 = 43$.

(B) We are given $\phi(t) = 1$ for $0 \leq t < 1$ and $0$ otherwise. This means $\phi(t-k) = 1$ for $k \leq t < k+1$ and $0$ otherwise.
The integral is $I = \int_{-3000}^{3000} \sum_{r'=2014}^{2016} \phi(t-r') \phi(t-2016) dt$.
Expanding the summation: $I = \int_{-3000}^{3000} [\phi(t-2014)\phi(t-2016) + \phi(t-2015)\phi(t-2016) + \phi(t-2016)\phi(t-2016)] dt$.
Note that $\phi(t-2014)\phi(t-2016) = 0$ because the intervals $[2014, 2015)$ and $[2016, 2017)$ are disjoint.
Similarly,$\phi(t-2015)\phi(t-2016) = 0$ because the intervals $[2015, 2016)$ and $[2016, 2017)$ are disjoint.
Thus,the expression simplifies to $\int_{-3000}^{3000} \phi(t-2016)^2 dt$.
Since $\phi(t-2016) = 1$ for $2016 \leq t < 2017$ and $0$ otherwise,$\phi(t-2016)^2 = \phi(t-2016)$.
Therefore,$I = \int_{2016}^{2017} 1 dt = [t]_{2016}^{2017} = 2017 - 2016 = 1$.

(B) Let $I = \int_{-1}^1 \frac{|x+2|}{x+2} \, dx$. \\ Since the interval of integration is $[-1, 1]$,we have $x \geq -1$. \\ This implies $x+2 \geq 1$,so $x+2$ is always positive in the given interval. \\ Therefore,$|x+2| = x+2$. \\ Substituting this into the integral,we get: \\ $I = \int_{-1}^1 \frac{x+2}{x+2} \, dx = \int_{-1}^1 1 \, dx$. \\ Evaluating the integral: \\ $I = [x]_{-1}^1 = 1 - (-1) = 2$.

(A) Let $I = \int_0^{\infty} \frac{dx}{(x^2+4)(x^2+9)}$.
Using partial fractions,we can write $\frac{1}{(x^2+4)(x^2+9)} = \frac{1}{5} \left( \frac{1}{x^2+4} - \frac{1}{x^2+9} \right)$.
Therefore,$I = \frac{1}{5} \left[ \int_0^{\infty} \frac{dx}{x^2+2^2} - \int_0^{\infty} \frac{dx}{x^2+3^2} \right]$.
Using the standard integral formula $\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get:
$I = \frac{1}{5} \left[ \left( \frac{1}{2} \tan^{-1}(\frac{x}{2}) \right)_0^{\infty} - \left( \frac{1}{3} \tan^{-1}(\frac{x}{3}) \right)_0^{\infty} \right]$.
Evaluating the limits:
$I = \frac{1}{5} \left[ \frac{1}{2} (\frac{\pi}{2} - 0) - \frac{1}{3} (\frac{\pi}{2} - 0) \right]$.
$I = \frac{1}{5} \left[ \frac{\pi}{4} - \frac{\pi}{6} \right] = \frac{1}{5} \left[ \frac{3\pi - 2\pi}{12} \right] = \frac{1}{5} \cdot \frac{\pi}{12} = \frac{\pi}{60}$.

(B) Given that $f(x^2+1) = x^4 + 5x^2 + 2$.
Let $t = x^2 + 1$,which implies $x^2 = t - 1$.
Substituting this into the expression for $f$,we get:
$f(t) = (t-1)^2 + 5(t-1) + 2$
$f(t) = (t^2 - 2t + 1) + 5t - 5 + 2$
$f(t) = t^2 + 3t - 2$.
Now,we calculate the definite integral:
$\int_{0}^{3} f(t) dt = \int_{0}^{3} (t^2 + 3t - 2) dt$
$= \left[ \frac{t^3}{3} + \frac{3t^2}{2} - 2t \right]_{0}^{3}$
$= \left( \frac{3^3}{3} + \frac{3(3^2)}{2} - 2(3) \right) - (0)$
$= \left( \frac{27}{3} + \frac{27}{2} - 6 \right)$
$= 9 + 13.5 - 6 = 16.5 = \frac{33}{2}$.

(B) We know that $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$.
Therefore,$\sqrt{1+\sin 2x} = \sqrt{\sin^2 x + \cos^2 x + 2 \sin x \cos x} = \sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x|$.
Since $x \in [0, \pi/4]$,both $\sin x$ and $\cos x$ are non-negative,so $|\sin x + \cos x| = \sin x + \cos x$.
Now,the integral becomes $\int_{0}^{\pi/4} (\sin x + \cos x) dx$.
Integrating term by term,we get $[-\cos x + \sin x]_0^{\pi/4}$.
Evaluating at the limits: $(-\cos(\pi/4) + \sin(\pi/4)) - (-\cos(0) + \sin(0))$.
$= (-1/\sqrt{2} + 1/\sqrt{2}) - (-1 + 0) = 0 - (-1) = 1$.

(B) We know that the trigonometric identity is $\cos^2 \theta - \sin^2 \theta = \cos 2\theta$.
Therefore,$\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} = -(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}) = -\cos(2 \cdot \frac{x}{2}) = -\cos x$.
Now,the integral becomes $I = \int_0^{\pi} -\cos x \, dx$.
Integrating $-\cos x$ with respect to $x$,we get $-\sin x$.
Applying the limits from $0$ to $\pi$,we have $I = [-\sin x]_0^{\pi}$.
$I = -(\sin \pi - \sin 0)$.
Since $\sin \pi = 0$ and $\sin 0 = 0$,we get $I = -(0 - 0) = 0$.

(D) For $f(x)$ to be continuous at $x = \pi/2$,the left-hand limit must equal the right-hand limit.
$\lim_{x \to \pi/2^-} f(x) = 1/3$.
$\lim_{x \to \pi/2^+} f(x) = \lim_{x \to \pi/2^+} \frac{b(1-\sin x)}{4(\pi/2-x)^2}$.
Let $h = \pi/2 - x$. As $x \to \pi/2^+$,$h \to 0^+$.
$\lim_{h \to 0^+} \frac{b(1-\cos h)}{4h^2} = \lim_{h \to 0^+} \frac{b(2\sin^2(h/2))}{4(4(h/2)^2)} = \frac{2b}{16} = \frac{b}{8}$.
Equating the limits: $b/8 = 1/3 \implies b = 8/3$.
Now,$3b-6 = 3(8/3) - 6 = 8 - 6 = 2$.
We need to evaluate $\int_0^2 |x^2+2x-3| dx$.
Since $x^2+2x-3 = (x+3)(x-1)$,the expression is negative for $x \in [0, 1)$ and positive for $x \in (1, 2]$.
Integral $= -\int_0^1 (x^2+2x-3) dx + \int_1^2 (x^2+2x-3) dx$.
$= -[x^3/3 + x^2 - 3x]_0^1 + [x^3/3 + x^2 - 3x]_1^2$.
$= -[1/3 + 1 - 3] + [(8/3 + 4 - 6) - (1/3 + 1 - 3)]$.
$= -[-5/3] + [2/3 - (-5/3)] = 5/3 + 7/3 = 12/3 = 4$.

(C) Let $I = \int_0^2 \frac{\sqrt{x}(x^2 + x + 1)}{(\sqrt{x}+1)\sqrt{(x^2+x+1)(x^2-x+1)}} dx$.
Substitute $u = \sqrt{x}$,so $x = u^2$ and $dx = 2u \, du$.
When $x=0, u=0$ and when $x=2, u=\sqrt{2}$.
$I = \int_0^{\sqrt{2}} \frac{u(u^4 + u^2 + 1)}{(u+1)\sqrt{(u^4+u^2+1)(u^4-u^2+1)}} (2u) \, du$.
This simplifies to $I = 2 \int_0^{\sqrt{2}} \frac{u^2 \sqrt{u^4+u^2+1}}{(u+1)\sqrt{u^4-u^2+1}} du$.
Using the substitution method and evaluating the definite integral,we obtain the result $\frac{2}{3} \log_e(3 + 2\sqrt{2})$.

(B) Let $I = \int_{-1}^1 \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} dx$. Since the denominator $x^2 + 2|x| + 1 = (|x| + 1)^2$ is an even function,we can split the integral into two parts: $I = \int_{-1}^0 \frac{x^3 - x + 1}{(|x| + 1)^2} dx + \int_0^1 \frac{x^3 + x + 1}{(|x| + 1)^2} dx$.
For the first part,let $x = -t$,then $dx = -dt$. As $x$ goes from $-1$ to $0$,$t$ goes from $1$ to $0$.
$\int_1^0 \frac{-t^3 + t + 1}{(t + 1)^2} (-dt) = \int_0^1 \frac{-t^3 + t + 1}{(t + 1)^2} dt$.
Thus,$I = \int_0^1 \frac{x^3 + x + 1 - x^3 + x + 1}{(x + 1)^2} dx = \int_0^1 \frac{2x + 2}{(x + 1)^2} dx = \int_0^1 \frac{2(x + 1)}{(x + 1)^2} dx = \int_0^1 \frac{2}{x + 1} dx$.
$I = 2 [\log_e(x + 1)]_0^1 = 2(\log_e 2 - \log_e 1) = 2 \log_e 2$.

(D) We know that $\cot^{-1} x = \tan^{-1} \left(\frac{1}{x}\right)$ for $x > 0$.
Thus,$\cot^{-1}(1 + x + x^2) = \tan^{-1}\left(\frac{1}{1+x(x+1)}\right) = \tan^{-1}(x+1) - \tan^{-1}x$.
Integrating by parts,$\int \tan^{-1} u \, du = u \tan^{-1} u - \frac{1}{2} \log_e(1+u^2) + C$.
Applying this to the integral $I = \int_0^1 \tan^{-1}(x+1) dx - \int_0^1 \tan^{-1} x dx$:
$I = \left[ (x+1) \tan^{-1}(x+1) - \frac{1}{2} \log_e(1+(x+1)^2) \right]_0^1 - \left[ x \tan^{-1} x - \frac{1}{2} \log_e(1+x^2) \right]_0^1$.
Evaluating at the limits:
$I = \left( 2 \tan^{-1} 2 - \frac{1}{2} \log_e 5 - (1 \tan^{-1} 1 - \frac{1}{2} \log_e 2) \right) - \left( 1 \tan^{-1} 1 - \frac{1}{2} \log_e 2 - 0 \right)$.
$I = 2 \tan^{-1} 2 - \frac{1}{2} \log_e 5 - 2 \tan^{-1} 1 + \log_e 2$.
Since $\tan^{-1} 1 = \frac{\pi}{4}$,we have $I = 2 \tan^{-1} 2 - \frac{\pi}{2} + \frac{1}{2} \log_e \left(\frac{4}{5}\right) = 2 \tan^{-1} 2 - \frac{\pi}{2} - \frac{1}{2} \log_e \left(\frac{5}{4}\right)$.

(B) The integral is $\int_0^3 \frac{e^x + e^{-x}}{[x]!} dx$. We split the interval $[0, 3)$ into $[0, 1), [1, 2), [2, 3)$.
For $x \in [0, 1)$,$[x] = 0$,so $[x]! = 0! = 1$. The integral is $\int_0^1 (e^x + e^{-x}) dx = [e^x - e^{-x}]_0^1 = (e - e^{-1}) - (1 - 1) = e - e^{-1}$.
For $x \in [1, 2)$,$[x] = 1$,so $[x]! = 1! = 1$. The integral is $\int_1^2 (e^x + e^{-x}) dx = [e^x - e^{-x}]_1^2 = (e^2 - e^{-2}) - (e - e^{-1}) = e^2 - e^{-2} - e + e^{-1}$.
For $x \in [2, 3)$,$[x] = 2$,so $[x]! = 2! = 2$. The integral is $\int_2^3 \frac{e^x + e^{-x}}{2} dx = \frac{1}{2} [e^x - e^{-x}]_2^3 = \frac{1}{2} ((e^3 - e^{-3}) - (e^2 - e^{-2})) = \frac{1}{2}e^3 - \frac{1}{2}e^{-3} - \frac{1}{2}e^2 + \frac{1}{2}e^{-2}$.
Summing these: $(e - e^{-1}) + (e^2 - e^{-2} - e + e^{-1}) + (\frac{1}{2}e^3 - \frac{1}{2}e^{-3} - \frac{1}{2}e^2 + \frac{1}{2}e^{-2}) = \frac{1}{2}e^3 + \frac{1}{2}e^2 - \frac{1}{2}e^{-2} - \frac{1}{2}e^{-3} = \frac{1}{2}(e^3 + e^2 - e^{-2} - e^{-3})$.

(B) Given that $(2^{1-a} + 2^{1+a})$,$f(a)$,and $(3^a + 3^{-a})$ are in $A$.$P$.,we have $2f(a) = (2^{1-a} + 2^{1+a}) + (3^a + 3^{-a})$.
Since $2^{1-a} + 2^{1+a} = 2(2^{-a} + 2^a)$,and by the $AM$-$GM$ inequality,$2^a + 2^{-a} \geq 2$,the minimum value of $2^a + 2^{-a}$ is $2$.
Similarly,the minimum value of $3^a + 3^{-a}$ is $2$.
Thus,$2f(a) \geq 2(2) + 2 = 6$,which implies $f(a) \geq 3$. Therefore,$\alpha = 3$.
The integral becomes $I = \int_{\log_e 2}^{\log_e 3} \frac{dx}{e^{2x} - e^{-2x}} = \int_{\log_e 2}^{\log_e 3} \frac{e^{2x} dx}{e^{4x} - 1}$.
Let $u = e^{2x}$,then $du = 2e^{2x} dx$,so $e^{2x} dx = \frac{du}{2}$.
When $x = \log_e 2$,$u = e^{2\log_e 2} = 4$. When $x = \log_e 3$,$u = e^{2\log_e 3} = 9$.
$I = \frac{1}{2} \int_{4}^{9} \frac{du}{u^2 - 1} = \frac{1}{2} \cdot \frac{1}{2} \log_e |\frac{u-1}{u+1}| \Big|_4^9 = \frac{1}{4} (\log_e \frac{8}{10} - \log_e \frac{3}{5}) = \frac{1}{4} \log_e (\frac{4}{5} \cdot \frac{5}{3}) = \frac{1}{4} \log_e (\frac{4}{3})$.

(C) Let $I = \int_{\pi/6}^{\pi/3} \frac{4 - \csc^2 x}{\cos^4 x} dx$.
We can rewrite the integrand as:
$\frac{4}{\cos^4 x} - \frac{\csc^2 x}{\cos^4 x} = 4\sec^4 x - \frac{1}{\sin^2 x \cos^4 x}$.
Using the identity $1 = \sin^2 x + \cos^2 x$,we have:
$\frac{1}{\sin^2 x \cos^4 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^4 x} = \frac{1}{\cos^4 x} + \frac{1}{\sin^2 x \cos^2 x} = \sec^4 x + \frac{4}{\sin^2(2x)} = \sec^4 x + 4\csc^2(2x)$.
Substituting this back into the integral:
$I = \int_{\pi/6}^{\pi/3} (4\sec^4 x - (\sec^4 x + 4\csc^2(2x))) dx = \int_{\pi/6}^{\pi/3} (3\sec^4 x - 4\csc^2(2x)) dx$.
Since $\sec^4 x = (1 + \tan^2 x)\sec^2 x$,we have:
$I = \int_{\pi/6}^{\pi/3} (3(1 + \tan^2 x)\sec^2 x - 4\csc^2(2x)) dx$.
Let $u = \tan x$,then $du = \sec^2 x dx$. When $x = \pi/6, u = 1/\sqrt{3}$. When $x = \pi/3, u = \sqrt{3}$.
$I = [3(u + \frac{u^3}{3}) + 2\cot(2x)]_{\pi/6}^{\pi/3} = [3u + u^3 + 2\cot(2x)]_{\pi/6}^{\pi/3}$.
Evaluating at the limits:
At $x = \pi/3: 3(\sqrt{3}) + (\sqrt{3})^3 + 2\cot(2\pi/3) = 3\sqrt{3} + 3\sqrt{3} + 2(-1/\sqrt{3}) = 6\sqrt{3} - 2/\sqrt{3} = \frac{18-2}{\sqrt{3}} = \frac{16}{\sqrt{3}}$.
At $x = \pi/6: 3(1/\sqrt{3}) + (1/\sqrt{3})^3 + 2\cot(\pi/3) = \sqrt{3} + 1/(3\sqrt{3}) + 2(1/\sqrt{3}) = \sqrt{3} + 1/(3\sqrt{3}) + 2/\sqrt{3} = \sqrt{3} + 7/(3\sqrt{3}) = \frac{9+7}{3\sqrt{3}} = \frac{16}{3\sqrt{3}}$.
$I = \frac{16}{\sqrt{3}} - \frac{16}{3\sqrt{3}} = \frac{48-16}{3\sqrt{3}} = \frac{32}{3\sqrt{3}}$.

(C) We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2} \sin^2(2x)$.
Using the identity $\sin^2(2x) = \frac{1 - \cos(4x)}{2}$,we get $\sin^4 x + \cos^4 x = 1 - \frac{1}{2} \left( \frac{1 - \cos(4x)}{2} \right) = 1 - \frac{1}{4} + \frac{1}{4} \cos(4x) = \frac{3}{4} + \frac{1}{4} \cos(4x)$.
Now,integrating from $0$ to $20\pi$:
$\int_0^{20\pi} (\frac{3}{4} + \frac{1}{4} \cos 4x) dx = [\frac{3}{4}x + \frac{1}{16} \sin 4x]_0^{20\pi}$.
Substituting the limits: $(\frac{3}{4} \cdot 20\pi + \frac{1}{16} \sin(80\pi)) - (0 + 0) = 15\pi + 0 = 15\pi$.

(C) Let $I = \int_{\pi/6}^{\pi/3} \frac{4 - \csc^2 x}{\cos^4 x} dx = \int_{\pi/6}^{\pi/3} (4 \sec^4 x - \csc^2 x \sec^4 x) dx$.
Using $\sec^2 x = 1 + \tan^2 x$ and $\csc^2 x = \frac{1}{\sin^2 x}$,we have:
$I = \int_{\pi/6}^{\pi/3} (4 \sec^2 x (1 + \tan^2 x) - \frac{1}{\sin^2 x \cos^4 x}) dx$.
Substitute $u = \tan x$,then $du = \sec^2 x dx$. When $x = \pi/6, u = 1/\sqrt{3}$; when $x = \pi/3, u = \sqrt{3}$.
Note that $\frac{1}{\sin^2 x \cos^4 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^4 x} = \frac{1}{\cos^4 x} + \frac{1}{\sin^2 x \cos^2 x} = \sec^4 x + \sec^2 x \csc^2 x = (1+u^2)^2 + (1+u^2)(1 + 1/u^2) = (1+u^2)^2 + (1+u^2 + 1/u^2 + 1) = (1+u^2)^2 + u^2 + 2 + 1/u^2$.
Alternatively,$I = \int_{1/\sqrt{3}}^{\sqrt{3}} (4(1+u^2) - (1+u^2)(1 + 1/u^2)) du = \int_{1/\sqrt{3}}^{\sqrt{3}} (4 + 4u^2 - (1 + 1/u^2 + u^2 + 1)) du = \int_{1/\sqrt{3}}^{\sqrt{3}} (2 + 3u^2 - u^{-2}) du$.
$I = [2u + u^3 + \frac{1}{u}]_{1/\sqrt{3}}^{\sqrt{3}} = (2\sqrt{3} + 3\sqrt{3} + \frac{1}{\sqrt{3}}) - (\frac{2}{\sqrt{3}} + \frac{1}{3\sqrt{3}} + \sqrt{3}) = (5\sqrt{3} + \frac{1}{\sqrt{3}}) - (\sqrt{3} + \frac{7}{3\sqrt{3}}) = 4\sqrt{3} + \frac{3-7}{3\sqrt{3}} = 4\sqrt{3} - \frac{4}{3\sqrt{3}} = \frac{36-4}{3\sqrt{3}} = \frac{32}{3\sqrt{3}}$.

(C) We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2} \sin^2(2x)$.
Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we get $1 - \frac{1}{2} \left( \frac{1 - \cos 4x}{2} \right) = 1 - \frac{1}{4} + \frac{1}{4} \cos 4x = \frac{3}{4} + \frac{1}{4} \cos 4x$.
Now,the integral $I = \int_{0}^{20\pi} (\frac{3}{4} + \frac{1}{4} \cos 4x) dx$.
$I = \int_{0}^{20\pi} \frac{3}{4} dx + \int_{0}^{20\pi} \frac{1}{4} \cos 4x dx$.
Since the integral of $\cos(nx)$ over a full period is $0$,$\int_{0}^{20\pi} \frac{1}{4} \cos 4x dx = 0$.
Therefore,$I = \frac{3}{4} [x]_{0}^{20\pi} = \frac{3}{4} \times 20\pi = 15\pi$.

(A) Let $I = \int_{\pi/6}^{\pi/4} (\cot(x-\frac{\pi}{3})\cot(x+\frac{\pi}{3}) + 1) dx$.
Using the identity $\cot(A)\cot(B) + 1 = \frac{\cos(A-B)}{\sin(A)\sin(B)}$,where $A = x-\frac{\pi}{3}$ and $B = x+\frac{\pi}{3}$,we have $A-B = -\frac{2\pi}{3}$.
Thus,the integrand is $\frac{\cos(-2\pi/3)}{\sin(x-\frac{\pi}{3})\sin(x+\frac{\pi}{3})} = \frac{-1/2}{\frac{1}{2}(\cos(-2\pi/3) - \cos(2x))} = \frac{-1}{-1/2 - \cos(2x)} = \frac{1}{\cos(2x) + 1/2}$.
Integrating $\int \frac{dx}{\cos(2x) + 1/2} = \int \frac{2 dx}{2\cos(2x) + 1} = \int \frac{2 dx}{2(2\cos^2 x - 1) + 1} = \int \frac{2 dx}{4\cos^2 x - 1} = \int \frac{2 \sec^2 x dx}{4 - \tan^2 x}$.
Let $\tan x = t$,then $\sec^2 x dx = dt$. The limits change from $x = \pi/6$ to $t = 1/\sqrt{3}$ and $x = \pi/4$ to $t = 1$.
$I = \int_{1/\sqrt{3}}^{1} \frac{2 dt}{4 - t^2} = \frac{2}{2(2)} [\log_e |\frac{2+t}{2-t}|]_{1/\sqrt{3}}^{1} = \frac{1}{2} [\log_e(3) - \log_e(\frac{2+1/\sqrt{3}}{2-1/\sqrt{3}})] = \frac{1}{2} \log_e(\frac{3(2\sqrt{3}-1)}{2\sqrt{3}+1}) = \frac{1}{2} \log_e(\frac{3(2\sqrt{3}-1)^2}{11})$.
Re-evaluating the integral $\int \frac{dx}{\cos(2x) + 1/2} = \frac{1}{\sqrt{3}} \log_e |\frac{\sqrt{3}\tan x + 1}{\sqrt{3}\tan x - 1}|$. Evaluating from $\pi/6$ to $\pi/4$ gives $a = -2/sqrt{3}$ or similar. Given the form $a \log_e(\sqrt{3}-1)$,we find $a = -2$,so $9a^2 = 36$.