Fundamental integration Questions in English

(B) We know that $1 + \sin x = \sin^2(x/2) + \cos^2(x/2) + 2 \sin(x/2) \cos(x/2) = (\sin(x/2) + \cos(x/2))^2$.
Thus, $\sqrt{1 + \sin x} = |\sin(x/2) + \cos(x/2)|$.
Assuming the interval where $\sin(x/2) + \cos(x/2) > 0$, we have $\int (\sin(x/2) + \cos(x/2)) \, dx$.
$= -2 \cos(x/2) + 2 \sin(x/2) = 2(\sin(x/2) - \cos(x/2))$.
Using $\sin \theta - \cos \theta = \sqrt{2} \sin(\theta - \pi/4) = -\sqrt{2} \cos(\theta + \pi/4)$.
So, $2(\sin(x/2) - \cos(x/2)) = -2\sqrt{2} \cos(x/2 + \pi/4)$.
Comparing this with $-4 \cos(ax + b)$, we note the coefficient mismatch. Let's re-evaluate: $\sqrt{1 + \sin x} = \sqrt{(\cos(x/2) + \sin(x/2))^2} = \cos(x/2) + \sin(x/2)$.
Integral $= 2 \sin(x/2) - 2 \cos(x/2) = 2\sqrt{2} (\frac{1}{\sqrt{2}} \sin(x/2) - \frac{1}{\sqrt{2}} \cos(x/2)) = 2\sqrt{2} \sin(x/2 - \pi/4) = -2\sqrt{2} \cos(x/2 + \pi/4)$.
Given the form $-4 \cos(ax+b)$, there is a constant factor discrepancy in the problem statement, but $a=1/2$ and $b=\pi/4$ match the argument of the cosine function.

(A) Step $1$: Use the identity $\cot^2 x = \csc^2 x - 1$.
$\int \cot^4 x \, dx = \int \cot^2 x (\csc^2 x - 1) \, dx$
Step $2$: Expand the integral.
$= \int \cot^2 x \csc^2 x \, dx - \int \cot^2 x \, dx$
Step $3$: Substitute $\cot^2 x = \csc^2 x - 1$ in the second integral.
$= \int \cot^2 x \csc^2 x \, dx - \int (\csc^2 x - 1) \, dx$
Step $4$: Integrate. For the first part, let $u = \cot x$, then $du = -\csc^2 x \, dx$.
$= -\frac{\cot^3 x}{3} - (-\cot x - x) + c$
$= -\frac{\cot^3 x}{3} + \cot x + x + c$

(B) The standard integral is $\int \frac{1}{Ax^2+B} dx = \frac{1}{\sqrt{AB}} \tan^{-1} \left(x \sqrt{\frac{A}{B}}\right) + c$.
Given $\int \frac{1}{ax^2+b} dx = \frac{1}{\sqrt{ab}} \tan^{-1} \left(x \sqrt{\frac{a}{b}}\right) + c = \frac{1}{\sqrt{6}} \tan^{-1} \left(x \sqrt{\frac{2}{3}}\right) + c$.
Comparing, we get $\sqrt{ab} = \sqrt{6} \implies ab = 6$ and $\sqrt{\frac{a}{b}} = \sqrt{\frac{2}{3}} \implies \frac{a}{b} = \frac{2}{3}$.
Solving $ab=6$ and $a = \frac{2b}{3}$, we get $(\frac{2b}{3})b = 6 \implies b^2 = 9 \implies b = 3$ and $a = 2$.
Now, $\int \frac{1}{bx^2+a} dx = \int \frac{1}{3x^2+2} dx = \frac{1}{\sqrt{3 \cdot 2}} \tan^{-1} \left(x \sqrt{\frac{3}{2}}\right) + c = \frac{1}{\sqrt{6}} \tan^{-1} \left(\frac{\sqrt{3}x}{\sqrt{2}}\right) + c$.

(D) Given $\int f'(x) e^{x^2} dx = (x - 1) e^{x^2} + k$.
Differentiating both sides with respect to $x$:
$f'(x) e^{x^2} = \frac{d}{dx} [(x - 1) e^{x^2}]$
$f'(x) e^{x^2} = (1) e^{x^2} + (x - 1) e^{x^2} (2x)$
$f'(x) e^{x^2} = e^{x^2} [1 + 2x^2 - 2x]$
$f'(x) = 2x^2 - 2x + 1$
Integrating $f'(x)$ with respect to $x$:
$f(x) = \int (2x^2 - 2x + 1) dx$
$f(x) = \frac{2x^3}{3} - x^2 + x + c$
Thus, option $D$ is correct.

(A) We know that $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$.
Let $a = \sin^2 x$ and $b = \cos^2 x$. Then $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)$.
Since $\sin^2 x + \cos^2 x = 1$, we have $\sin^6 x + \cos^6 x = \sin^4 x - \sin^2 x \cos^2 x + \cos^4 x$.
Adding and subtracting $2 \sin^2 x \cos^2 x$, we get $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)^2 - 3 \sin^2 x \cos^2 x = 1 - 3 \sin^2 x \cos^2 x$.
Now, the integral becomes $\int \frac{1 - 3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} dx = \int (\frac{1}{\sin^2 x \cos^2 x} - 3) dx$.
Using $1 = (\sin^2 x + \cos^2 x)^2$, we have $\int (\frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} - 3) dx = \int (\sec^2 x + \csc^2 x - 3) dx$.
Integrating term by term, we get $\tan x - \cot x - 3x + c$.

(D) Consider the derivative of the quotient $\frac{u^2}{2v^2}$ with respect to $x$ using the quotient rule:
$\frac{d}{dx} (\frac{u^2}{2v^2}) = \frac{1}{2} \cdot \frac{v^2 \frac{d}{dx}(u^2) - u^2 \frac{d}{dx}(v^2)}{(v^2)^2}$
$= \frac{1}{2} \cdot \frac{v^2 (2u \frac{du}{dx}) - u^2 (2v \frac{dv}{dx})}{v^4}$
$= \frac{v^2 u \frac{du}{dx} - u^2 v \frac{dv}{dx}}{v^4}$
$= \frac{v(uv \frac{du}{dx} - u^2 \frac{dv}{dx})}{v^4}$
$= \frac{uv \frac{du}{dx} - u^2 \frac{dv}{dx}}{v^3}$
Since the derivative of $\frac{u^2}{2v^2}$ is the integrand, the integral is $\frac{u^2}{2v^2} + c$.

(A) Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:
$\sin 4x \cos 3x = \frac{1}{2} [\sin(4x+3x) + \sin(4x-3x)] = \frac{1}{2} [\sin 7x + \sin x]$
Now, integrate the expression:
$\int \sin 4x \cos 3x \, dx = \frac{1}{2} \int (\sin 7x + \sin x) \, dx$
$= \frac{1}{2} [-\frac{\cos 7x}{7} - \cos x] + c$
$= -\frac{1}{14} \cos 7x - \frac{1}{2} \cos x + c$

(B) Step $1$: Simplify the integrand using the property $e^{\log a} = a$. We have $e^{-x \log 2} = (e^{\log 2})^{-x} = 2^{-x}$.
Step $2$: Substitute this into the integral: $\int 2^{-x} \cdot 2^x dx$.
Step $3$: Simplify the product: $\int 2^{-x+x} dx = \int 2^0 dx = \int 1 dx$.
Step $4$: Integrate with respect to $x$: $\int 1 dx = x + C$.