Fundamental integration Questions in English

(A) The integral is with respect to the variable $a$.
Since $x$ is treated as a constant,we use the power rule for integration: $\int a^n \, da = \frac{a^{n+1}}{n+1} + c$.
Substituting $n = x$,we get:
$\int a^x \, da = \frac{a^{x+1}}{x+1} + c$.
Therefore,the correct option is $A$.

(C) We know that $\cot x = \frac{\cos x}{\sin x}$.
Therefore,$\int \cot x \, dx = \int \frac{\cos x}{\sin x} \, dx$.
Let $u = \sin x$,then $du = \cos x \, dx$.
Substituting these into the integral,we get $\int \frac{1}{u} \, du = \log |u| + c$.
Substituting back $u = \sin x$,we obtain $\log |\sin x| + c$.

(A) To find the integral $\int \frac{1}{x^4} \, dx$,we rewrite the integrand as a power of $x$:
$\int x^{-4} \, dx$.
Using the power rule for integration,$\int x^n \, dx = \frac{x^{n+1}}{n+1} + c$ (where $n \neq -1$):
$\int x^{-4} \, dx = \frac{x^{-4+1}}{-4+1} + c = \frac{x^{-3}}{-3} + c$.
Simplifying the expression,we get:
$-\frac{1}{3x^3} + c$.

(C) We know that ${e^{\log f(x)}} = f(x)$.
Therefore,the integral becomes:
$\int {\frac{{{x^5} - {x^4}}}{{{x^3} - {x^2}}}\,dx}$
$= \int {\frac{{{x^4}(x - 1)}}{{{x^2}(x - 1)}}\,dx}$
$= \int {{x^2}\,dx}$
$= \frac{{{x^3}}}{3} + c$.

(A) To solve the integral $\int {\frac{{{x^4} + {x^2} + 1}}{{{x^2} - x + 1}}\,dx}$,we first simplify the integrand by factoring the numerator.
We know that ${x^4} + {x^2} + 1 = ({x^2} + 1)^2 - {x^2} = ({x^2} + 1 - x)({x^2} + 1 + x)$.
Substituting this into the integral,we get:
$\int {\frac{({x^2} - x + 1)({x^2} + x + 1)}{{{x^2} - x + 1}}\,dx} = \int {({x^2} + x + 1)\,dx}$.
Now,integrate term by term:
$\int {x^2\,dx} + \int {x\,dx} + \int {1\,dx} = \frac{{{x^3}}}{3} + \frac{{{x^2}}}{2} + x + c$.
Thus,the correct option is $A$.

(B) The integral of $\sec x$ is given by:
$\int \sec x \, dx = \log |\sec x + \tan x| + c$
We know that $\sec x + \tan x = \frac{1}{\sec x - \tan x}$.
Substituting this into the expression:
$\int \sec x \, dx = \log \left| \frac{1}{\sec x - \tan x} \right| + c$
Using the property $\log(a^{-1}) = -\log(a)$:
$\int \sec x \, dx = -\log |\sec x - \tan x| + c$.
Thus,option $B$ is correct.

(D) We know that $1 + \sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$.
Thus,$\int \sqrt{1 + \sin x} \, dx = \int |\cos \frac{x}{2} + \sin \frac{x}{2}| \, dx$.
Assuming the interval where $\cos \frac{x}{2} + \sin \frac{x}{2} > 0$,we have:
$\int (\cos \frac{x}{2} + \sin \frac{x}{2}) \, dx = 2 \sin \frac{x}{2} - 2 \cos \frac{x}{2} + c = -2(\cos \frac{x}{2} - \sin \frac{x}{2}) + c$.
Since $(\cos \frac{x}{2} - \sin \frac{x}{2})^2 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2} = 1 - \sin x$,we can write the result as $-2\sqrt{1 - \sin x} + c$.

(B) We know that the derivative of $\cot x$ with respect to $x$ is $-\csc^2 x$.
Therefore,by the definition of the indefinite integral,$\int \csc^2 x \, dx = -\cot x + c$,where $c$ is the constant of integration.

(A) To evaluate the integral $\int (2\sin x + \frac{1}{x}) \, dx$,we use the linearity property of integration:
$\int (2\sin x + \frac{1}{x}) \, dx = \int 2\sin x \, dx + \int \frac{1}{x} \, dx$
We know that $\int \sin x \, dx = -\cos x + c_1$ and $\int \frac{1}{x} \, dx = \log |x| + c_2$.
Therefore,$\int 2\sin x \, dx + \int \frac{1}{x} \, dx = 2(-\cos x) + \log |x| + c = -2\cos x + \log |x| + c$.
Thus,the correct option is $A$.

(A) We know that $1 + \cos x = 2 \cos^2 \frac{x}{2}$.
Substituting this into the integral,we get:
$I = \int \sqrt{2 \cos^2 \frac{x}{2}} \, dx$
$I = \sqrt{2} \int \cos \frac{x}{2} \, dx$
Using the integration formula $\int \cos(ax) \, dx = \frac{1}{a} \sin(ax) + c$,where $a = \frac{1}{2}$:
$I = \sqrt{2} \cdot \frac{\sin(x/2)}{1/2} + c$
$I = 2\sqrt{2} \sin \frac{x}{2} + c$.

(D) We know that $2\sin x \cos x = \sin 2x$.
Substituting this into the integral,we get:
$I = \int \sin 2x \,dx$
Using the integration formula $\int \sin(ax) \,dx = -\frac{\cos(ax)}{a} + c$,we have:
$I = -\frac{\cos 2x}{2} + c$
Alternatively,using the identity $\cos 2x = 1 - 2\sin^2 x$:
$I = -\frac{1 - 2\sin^2 x}{2} + c$
$I = -\frac{1}{2} + \sin^2 x + c$
Since $c$ is an arbitrary constant,$- \frac{1}{2} + c$ can be written as a new constant $C$.
Thus,$I = \sin^2 x + C$.
Therefore,the correct option is $(d)$.

(B) To evaluate the integral $\int \tan^2 x \, dx$,we use the trigonometric identity $\tan^2 x = \sec^2 x - 1$.
Substituting this into the integral,we get:
$\int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx$
Using the linearity property of integration,we split the integral:
$= \int \sec^2 x \, dx - \int 1 \, dx$
We know that $\int \sec^2 x \, dx = \tan x$ and $\int 1 \, dx = x$.
Therefore,the final result is $\tan x - x + c$.

(A) We know that $e^{\log(f(x))} = f(x)$.
Therefore,the given integral becomes $\int {e^{\log (\sin x)}} \, dx = \int \sin x \, dx$.
The integral of $\sin x$ with respect to $x$ is $-\cos x + c$.
Thus,$\int \sin x \, dx = -\cos x + c$.

(B) We are given the integral $I = \int e^{x \log a} \cdot e^x \, dx$.
Using the property of logarithms,$e^{x \log a} = e^{\log(a^x)} = a^x$.
Substituting this into the integral,we get $I = \int a^x \cdot e^x \, dx$.
This simplifies to $I = \int (ae)^x \, dx$.
Using the standard integration formula $\int k^x \, dx = \frac{k^x}{\log k} + C$,where $k = ae$,we get $I = \frac{(ae)^x}{\log(ae)} + C$.

(A) We know that $1 + \cos x = 2 \cos^2 \frac{x}{2}$.
Substituting this into the integral:
$\int \frac{1}{\sqrt{2 \cos^2 \frac{x}{2}}} \, dx = \int \frac{1}{\sqrt{2} \cos \frac{x}{2}} \, dx$
$= \frac{1}{\sqrt{2}} \int \sec \frac{x}{2} \, dx$
Using the formula $\int \sec \theta \, d\theta = \log |\sec \theta + \tan \theta| + C$ and applying the chain rule for $\frac{x}{2}$ (where the derivative is $\frac{1}{2}$):
$= \frac{1}{\sqrt{2}} \cdot \frac{\log |\sec \frac{x}{2} + \tan \frac{x}{2}|}{1/2} + K$
$= \frac{2}{\sqrt{2}} \log |\sec \frac{x}{2} + \tan \frac{x}{2}| + K$
$= \sqrt{2} \log |\sec \frac{x}{2} + \tan \frac{x}{2}| + K$.

(C) Let $I = \int \frac{\cos 2x - 1}{\cos 2x + 1} dx$.
Using the trigonometric identities $\cos 2x = 1 - 2\sin^2 x$ and $\cos 2x = 2\cos^2 x - 1$,we have:
$I = \int \frac{(1 - 2\sin^2 x) - 1}{(2\cos^2 x - 1) + 1} dx$
$I = \int \frac{-2\sin^2 x}{2\cos^2 x} dx$
$I = - \int \tan^2 x dx$
Using the identity $\tan^2 x = \sec^2 x - 1$:
$I = - \int (\sec^2 x - 1) dx$
$I = - \int \sec^2 x dx + \int 1 dx$
$I = - \tan x + x + c$
$I = x - \tan x + c$

(C) Given integral: $I = \int {\frac{{a{x^3} + b{x^2} + c}}{{{x^4}}}dx}$
Divide each term in the numerator by the denominator:
$I = \int {\left( {\frac{{a{x^3}}}{{{x^4}}} + \frac{{b{x^2}}}{{{x^4}}} + \frac{c}{{{x^4}}}} \right)dx}$
$I = \int {\left( {\frac{a}{x} + \frac{b}{{{x^2}}} + \frac{c}{{{x^4}}}} \right)dx}$
$I = \int {\frac{a}{x}dx + \int {b{x^{ - 2}}dx + \int {c{x^{ - 4}}dx} } }$
Using the integration formulas $\int {\frac{1}{x}dx = \log |x|}$ and $\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}}}$:
$I = a\log |x| + b\left( {\frac{{{x^{ - 1}}}}{{ - 1}}} \right) + c\left( {\frac{{{x^{ - 3}}}}{{ - 3}}} \right) + C$
$I = a\log |x| - \frac{b}{x} - \frac{c}{{3{x^3}}} + C$
Thus,the correct option is $C$.

(B) Let $I = \int {\frac{1}{{{{(x - 5)}^2}}}dx} $.
We can rewrite the integrand as $(x - 5)^{-2}$.
Using the power rule for integration,$\int {x^n dx} = \frac{x^{n+1}}{n+1} + c$ (where $n \neq -1$):
$I = \int {(x - 5)^{-2} dx} = \frac{{{{(x - 5)}^{ - 2 + 1}}}}{{ - 2 + 1}} + c$
$I = \frac{{{{(x - 5)}^{ - 1}}}}{{ - 1}} + c$
$I = - \frac{1}{{(x - 5)}} + c$.

(A) Given integral: $I = \int \sqrt{2} \sqrt{1 + \sin x} \, dx$.
Using the identity $1 + \sin x = \sin^2(x/2) + \cos^2(x/2) + 2 \sin(x/2) \cos(x/2) = (\sin(x/2) + \cos(x/2))^2$.
So,$I = \int \sqrt{2} (\sin(x/2) + \cos(x/2)) \, dx$.
$I = \sqrt{2} \int \sin(x/2) \, dx + \sqrt{2} \int \cos(x/2) \, dx$.
$I = \sqrt{2} [ -2 \cos(x/2) + 2 \sin(x/2) ] + c$.
$I = 2\sqrt{2} [ \sin(x/2) - \cos(x/2) ] + c$.
Multiplying and dividing by $\sqrt{2}$:
$I = 2\sqrt{2} \cdot \sqrt{2} [ \frac{1}{\sqrt{2}} \sin(x/2) - \frac{1}{\sqrt{2}} \cos(x/2) ] + c$.
$I = 4 [ \sin(x/2) \cos(\pi/4) - \cos(x/2) \sin(\pi/4) ] + c$.
$I = 4 \sin(x/2 - \pi/4) + c$.
Since $\sin(\theta) = -\cos(\theta + \pi/2)$,we have:
$I = -4 \cos(x/2 - \pi/4 + \pi/2) + c = -4 \cos(x/2 + \pi/4) + c$.
Comparing with $-4 \cos(ax + b) + c$,we get $a = 1/2$ and $b = \pi/4$.

(A) The integral of an exponential function of the form $\int a^x \, dx$ is given by the formula $\int a^x \, dx = \frac{a^x}{\ln a} + c$,where $a > 0$ and $a \neq 1$.
Here,$a = 13$.
Therefore,$\int 13^x \, dx = \frac{13^x}{\ln 13} + c$.
Since $\log 13$ typically denotes the natural logarithm in calculus contexts,the correct option is $A$.

(A) We need to evaluate the integral $I = \int a^x \, dx$.
Recall the derivative of the exponential function $a^x$ with respect to $x$:
$\frac{d}{dx}(a^x) = a^x \log a$.
Dividing both sides by $\log a$ (where $a > 0$ and $a \neq 1$),we get:
$\frac{d}{dx} \left( \frac{a^x}{\log a} \right) = a^x$.
By the definition of the indefinite integral,if $\frac{d}{dx} F(x) = f(x)$,then $\int f(x) \, dx = F(x) + c$.
Therefore,$\int a^x \, dx = \frac{a^x}{\log a} + c$.

(B) We know that the derivative of $\sec x$ with respect to $x$ is given by:
$\frac{d}{dx}(\sec x) = \sec x \tan x$
By the definition of the indefinite integral (antiderivative),if $\frac{d}{dx}(F(x)) = f(x)$,then $\int f(x) \, dx = F(x) + c$.
Therefore,$\int \sec x \tan x \, dx = \sec x + c$.
Thus,the correct option is $B$.

(B) We are given the integral $I = \int (\sin^4 x - \cos^4 x) \, dx$.
Using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$,we can write:
$\sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)$.
Since $\sin^2 x + \cos^2 x = 1$,the expression simplifies to:
$\sin^4 x - \cos^4 x = \sin^2 x - \cos^2 x = -(\cos^2 x - \sin^2 x)$.
Using the trigonometric identity $\cos 2x = \cos^2 x - \sin^2 x$,we get:
$\sin^4 x - \cos^4 x = -\cos 2x$.
Now,substitute this into the integral:
$I = \int -\cos 2x \, dx = -\frac{\sin 2x}{2} + c$.

(B) We are given the integral $I = \int \frac{(x + 1)^2}{x(x^2 + 1)} \, dx$.
Expanding the numerator,we get $(x + 1)^2 = x^2 + 1 + 2x$.
Substituting this into the integral,we have $I = \int \frac{x^2 + 1 + 2x}{x(x^2 + 1)} \, dx$.
Splitting the integral,we get $I = \int \frac{x^2 + 1}{x(x^2 + 1)} \, dx + \int \frac{2x}{x(x^2 + 1)} \, dx$.
Simplifying the terms,we get $I = \int \frac{1}{x} \, dx + 2 \int \frac{1}{x^2 + 1} \, dx$.
Integrating these standard forms,we obtain $I = \log_e |x| + 2\tan^{-1} x + c$.

(B) Let $I = \int \frac{dx}{\sqrt{1 - x}}$.
We can rewrite the integral as $I = \int (1 - x)^{-1/2} dx$.
Using the integration formula $\int (ax + b)^n dx = \frac{(ax + b)^{n+1}}{a(n+1)} + c$,where $a = -1$,$b = 1$,and $n = -1/2$:
$I = \frac{(1 - x)^{-1/2 + 1}}{(-1)(-1/2 + 1)} + c$
$I = \frac{(1 - x)^{1/2}}{(-1)(1/2)} + c$
$I = \frac{\sqrt{1 - x}}{-1/2} + c$
$I = -2\sqrt{1 - x} + c$.

(C) We need to evaluate the integral $I = \int \frac{dx}{4x^2 + 9}$.
Factor out $4$ from the denominator:
$I = \int \frac{dx}{4(x^2 + \frac{9}{4})} = \frac{1}{4} \int \frac{dx}{x^2 + (\frac{3}{2})^2}$.
Using the standard integral formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$,where $a = \frac{3}{2}$:
$I = \frac{1}{4} \cdot \frac{1}{3/2} \tan^{-1}(\frac{x}{3/2}) + c$.
$I = \frac{1}{4} \cdot \frac{2}{3} \tan^{-1}(\frac{2x}{3}) + c$.
$I = \frac{1}{6} \tan^{-1}(\frac{2x}{3}) + c$.

(B) Let $I = \int \frac{dx}{\sqrt{x^2 - a^2}}$.
Substitute $x = a \sec \theta$,then $dx = a \sec \theta \tan \theta \, d\theta$.
Substituting these into the integral:
$I = \int \frac{a \sec \theta \tan \theta \, d\theta}{\sqrt{a^2 \sec^2 \theta - a^2}} = \int \frac{a \sec \theta \tan \theta \, d\theta}{a \tan \theta} = \int \sec \theta \, d\theta$.
The integral of $\sec \theta$ is $\log_e |\sec \theta + \tan \theta| + c$.
Since $x = a \sec \theta$,we have $\sec \theta = \frac{x}{a}$ and $\tan \theta = \sqrt{\sec^2 \theta - 1} = \sqrt{\frac{x^2}{a^2} - 1} = \frac{\sqrt{x^2 - a^2}}{a}$.
Therefore,$I = \log_e \left| \frac{x}{a} + \frac{\sqrt{x^2 - a^2}}{a} \right| + c = \log_e \left| \frac{x + \sqrt{x^2 - a^2}}{a} \right| + c$.
Using the property $\log(m/n) = \log m - \log n$,we get $\log_e |x + \sqrt{x^2 - a^2}| - \log_e |a| + c$. Since $-\log_e |a| + c$ is just another constant,we write it as $c$.
Thus,$I = \log_e |x + \sqrt{x^2 - a^2}| + c$.

(A) Let $I = \int \frac{x^2}{x^2 + 4} dx$.
We can rewrite the integrand as:
$I = \int \frac{x^2 + 4 - 4}{x^2 + 4} dx$
$I = \int \left( 1 - \frac{4}{x^2 + 4} \right) dx$
$I = \int 1 dx - 4 \int \frac{1}{x^2 + 2^2} dx$
Using the standard integral formula $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$:
$I = x - 4 \left( \frac{1}{2} \tan^{-1}(\frac{x}{2}) \right) + c$
$I = x - 2 \tan^{-1}(\frac{x}{2}) + c$.

(C) To evaluate the integral $I = \int \frac{dx}{a^2 - x^2}$,we use partial fraction decomposition.
We can write the integrand as:
$\frac{1}{a^2 - x^2} = \frac{1}{(a - x)(a + x)} = \frac{1}{2a} \left( \frac{1}{a + x} + \frac{1}{a - x} \right)$.
Now,integrating both sides with respect to $x$:
$I = \frac{1}{2a} \int \left( \frac{1}{a + x} + \frac{1}{a - x} \right) dx$
$I = \frac{1}{2a} \left( \int \frac{1}{a + x} dx + \int \frac{1}{a - x} dx \right)$
$I = \frac{1}{2a} \left( \log |a + x| - \log |a - x| \right) + C$
Using the logarithmic property $\log m - \log n = \log \left( \frac{m}{n} \right)$:
$I = \frac{1}{2a} \log \left| \frac{a + x}{a - x} \right| + C$.
Thus,the correct option is $C$.

(C) The integral $\int \frac{dx}{\sqrt{x^2 + a^2}}$ is a standard integral formula.
By using the substitution $x = a \tan \theta$,we have $dx = a \sec^2 \theta \ d\theta$.
Then,$\sqrt{x^2 + a^2} = \sqrt{a^2 \tan^2 \theta + a^2} = a \sec \theta$.
The integral becomes $\int \frac{a \sec^2 \theta \ d\theta}{a \sec \theta} = \int \sec \theta \ d\theta$.
The integral of $\sec \theta$ is $\log |\sec \theta + \tan \theta| + c$.
Substituting back $\tan \theta = \frac{x}{a}$ and $\sec \theta = \frac{\sqrt{x^2 + a^2}}{a}$,we get $\log |\frac{\sqrt{x^2 + a^2}}{a} + \frac{x}{a}| + c = \log |\frac{x + \sqrt{x^2 + a^2}}{a}| + c = \log |x + \sqrt{x^2 + a^2}| - \log |a| + c$.
Since $-\log |a| + c$ is a constant,we write the final result as $\log |x + \sqrt{x^2 + a^2}| + c$.

(A) Let $I = \int \frac{x - 2}{x^2 - 4x + 3} dx$.
Substitute $t = x^2 - 4x + 3$.
Then $dt = (2x - 4) dx = 2(x - 2) dx$,which implies $(x - 2) dx = \frac{1}{2} dt$.
Substituting these into the integral:
$I = \int \frac{1}{2t} dt = \frac{1}{2} \log |t| + c$.
Substituting back $t = x^2 - 4x + 3$:
$I = \frac{1}{2} \log |x^2 - 4x + 3| + c = \log \sqrt{|x^2 - 4x + 3|} + c$.
Thus,the correct option is $A$.

(C) We need to evaluate the integral $I = \int \frac{x + 1}{\sqrt{x^2 + 1}} dx$.
Split the integral into two parts: $I = \int \frac{x}{\sqrt{x^2 + 1}} dx + \int \frac{1}{\sqrt{x^2 + 1}} dx$.
For the first part,let $t = x^2 + 1$,then $dt = 2x dx$,which implies $x dx = \frac{1}{2} dt$.
So,$\int \frac{x}{\sqrt{x^2 + 1}} dx = \frac{1}{2} \int t^{-1/2} dt = \frac{1}{2} \cdot \frac{t^{1/2}}{1/2} = \sqrt{t} = \sqrt{x^2 + 1}$.
For the second part,we use the standard integral formula $\int \frac{1}{\sqrt{x^2 + a^2}} dx = \log |x + \sqrt{x^2 + a^2}| + c$.
Thus,$\int \frac{1}{\sqrt{x^2 + 1}} dx = \log |x + \sqrt{x^2 + 1}| + c$.
Combining both parts,we get $I = \sqrt{x^2 + 1} + \log |x + \sqrt{x^2 + 1}| + c$.

(B) Let $t = 3x - 5$. Then,$dt = 3 \, dx$,which implies $dx = \frac{1}{3} \, dt$.
Substituting these into the integral:
$\int \tan(3x - 5) \sec(3x - 5) \, dx = \int \tan(t) \sec(t) \cdot \frac{1}{3} \, dt$
$= \frac{1}{3} \int \sec(t) \tan(t) \, dt$
Since the integral of $\sec(t) \tan(t)$ is $\sec(t) + c$,we get:
$= \frac{1}{3} \sec(t) + c$
Substituting back $t = 3x - 5$:
$= \frac{1}{3} \sec(3x - 5) + c$.

(B) To evaluate the integral $\int \tan^4 x \, dx$,we use the identity $\tan^2 x = \sec^2 x - 1$.
$\int \tan^4 x \, dx = \int \tan^2 x (\tan^2 x) \, dx$
$= \int \tan^2 x (\sec^2 x - 1) \, dx$
$= \int \tan^2 x \sec^2 x \, dx - \int \tan^2 x \, dx$
$= \int \tan^2 x \sec^2 x \, dx - \int (\sec^2 x - 1) \, dx$
For the first part,let $u = \tan x$,then $du = \sec^2 x \, dx$.
$\int u^2 \, du = \frac{u^3}{3} = \frac{\tan^3 x}{3}$.
For the second part,$\int (\sec^2 x - 1) \, dx = \tan x - x$.
Combining these,we get:
$\frac{1}{3} \tan^3 x - (\tan x - x) + c = \frac{1}{3} \tan^3 x - \tan x + x + c$.

(A) Let $f(x) = t$. Then,$f'(x) \, dx = dt$.
Substituting these into the integral,we get:
$\int \frac{f'(x)}{[f(x)]^2} \, dx = \int \frac{1}{t^2} \, dt$
$= \int t^{-2} \, dt$
$= \frac{t^{-2+1}}{-2+1} + c$
$= \frac{t^{-1}}{-1} + c$
$= -\frac{1}{t} + c$
$= -\frac{1}{f(x)} + c$
$= -[f(x)]^{-1} + c$.

(C) We have the integral $I = \int \frac{\sin 2x}{\sin 5x \sin 3x} \, dx$.
Since $2x = 5x - 3x$,we can write $\sin 2x = \sin(5x - 3x)$.
Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we get:
$I = \int \frac{\sin 5x \cos 3x - \cos 5x \sin 3x}{\sin 5x \sin 3x} \, dx$
$I = \int \left( \frac{\sin 5x \cos 3x}{\sin 5x \sin 3x} - \frac{\cos 5x \sin 3x}{\sin 5x \sin 3x} \right) \, dx$
$I = \int \left( \cot 3x - \cot 5x \right) \, dx$
Using the formula $\int \cot(ax) \, dx = \frac{1}{a} \log |\sin(ax)| + c$,we get:
$I = \frac{1}{3} \log |\sin 3x| - \frac{1}{5} \log |\sin 5x| + c$.

(B) Let $I = \int \frac{a^{\sqrt{x}}}{\sqrt{x}} dx$.
Substitute $\sqrt{x} = t$.
Differentiating both sides with respect to $x$,we get $\frac{1}{2\sqrt{x}} dx = dt$,which implies $\frac{dx}{\sqrt{x}} = 2 dt$.
Substituting these into the integral:
$I = \int a^t (2 dt) = 2 \int a^t dt$.
Using the standard integral formula $\int a^t dt = \frac{a^t}{\log_e a} + c$,we get:
$I = 2 \left( \frac{a^t}{\log_e a} \right) + c$.
Since $\frac{1}{\log_e a} = \log_a e$,we have:
$I = 2 a^t \log_a e + c$.
Substituting $t = \sqrt{x}$ back,we get:
$I = 2 a^{\sqrt{x}} \log_a e + c$.

(B) To evaluate the integral $I = \int a^{3x + 3} dx$,we use the method of substitution.
Let $t = 3x + 3$.
Then,differentiating both sides with respect to $x$,we get $dt = 3 dx$,which implies $dx = \frac{1}{3} dt$.
Substituting these into the integral:
$I = \int a^t \cdot \frac{1}{3} dt = \frac{1}{3} \int a^t dt$.
Using the standard integration formula $\int a^t dt = \frac{a^t}{\log_e a} + c$,we get:
$I = \frac{1}{3} \cdot \frac{a^t}{\log_e a} + c$.
Finally,substituting $t = 3x + 3$ back into the expression:
$I = \frac{a^{3x + 3}}{3 \log_e a} + c$.

(A) Let $I = \int \frac{1 + x^2}{\sqrt{1 - x^2}} dx$.
Substitute $x = \sin \theta$,so $dx = \cos \theta d\theta$.
The integral becomes $I = \int \frac{1 + \sin^2 \theta}{\cos \theta} \cos \theta d\theta = \int (1 + \sin^2 \theta) d\theta$.
Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we get $I = \int (1 + \frac{1 - \cos 2\theta}{2}) d\theta = \int (\frac{3}{2} - \frac{1}{2} \cos 2\theta) d\theta$.
Integrating,we get $I = \frac{3}{2} \theta - \frac{1}{4} \sin 2\theta + c = \frac{3}{2} \theta - \frac{1}{2} \sin \theta \cos \theta + c$.
Since $\theta = \sin^{-1} x$ and $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - x^2}$,we have $I = \frac{3}{2} \sin^{-1} x - \frac{1}{2} x \sqrt{1 - x^2} + c$.

(C) Given $I = \int \frac{\sin x}{\cos^2 x} \, dx$.
We can rewrite the integral as $I = \int \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} \, dx = \int \sec x \tan x \, dx$.
We know that the derivative of $\sec x$ is $\sec x \tan x$.
Therefore,$\int \sec x \tan x \, dx = \sec x + k$,where $k$ is the constant of integration.

(A) Let $I = \int \frac{dx}{x\sqrt{x^4 - 1}}$.
Multiply the numerator and denominator by $x$:
$I = \int \frac{x \, dx}{x^2 \sqrt{(x^2)^2 - 1}}$.
Substitute $t = x^2$,then $dt = 2x \, dx$,which implies $x \, dx = \frac{dt}{2}$.
Substituting these into the integral:
$I = \int \frac{dt/2}{t \sqrt{t^2 - 1}} = \frac{1}{2} \int \frac{dt}{t \sqrt{t^2 - 1}}$.
Using the standard integral formula $\int \frac{dt}{t \sqrt{t^2 - 1}} = \sec^{-1}(t) + k$:
$I = \frac{1}{2} \sec^{-1}(t) + k$.
Substituting back $t = x^2$:
$I = \frac{1}{2} \sec^{-1}(x^2) + k$.

(C) We need to evaluate the integral $I = \int \sin^3 x \, dx$.
Using the identity $\sin^2 x = 1 - \cos^2 x$,we can rewrite the integral as:
$I = \int \sin^2 x \cdot \sin x \, dx = \int (1 - \cos^2 x) \sin x \, dx$.
Let $u = \cos x$,then $du = -\sin x \, dx$,which implies $\sin x \, dx = -du$.
Substituting these into the integral:
$I = \int (1 - u^2) (-du) = \int (u^2 - 1) \, du$.
Integrating with respect to $u$:
$I = \frac{u^3}{3} - u + C$.
Substituting $u = \cos x$ back into the expression:
$I = \frac{\cos^3 x}{3} - \cos x + C$.

(A) Let $I = \int {{x^x}(1 + \log x)\,dx} $.
Consider the function $f(x) = x^x$.
Taking the natural logarithm on both sides,we get $\log f(x) = x \log x$.
Differentiating both sides with respect to $x$,we get $\frac{1}{f(x)} f'(x) = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$.
Thus,$f'(x) = f(x)(1 + \log x) = x^x(1 + \log x)$.
Since the derivative of $x^x$ is $x^x(1 + \log x)$,the integral of $x^x(1 + \log x)$ is $x^x + C$.

(D) Let $I = \int {\frac{{1 + {{\tan }^2}x}}{{1 - {{\tan }^2}x}}dx} $.
Using the identity $1 + \tan^2 x = \sec^2 x$,we get $I = \int {\frac{{{{\sec }^2}x}}{{1 - {{\tan }^2}x}}dx} $.
Substitute $\tan x = t$,then $\sec^2 x \, dx = dt$.
The integral becomes $I = \int {\frac{{dt}}{{1 - {t^2}}}} $.
Using the standard formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + c$,we have:
$I = \frac{1}{2(1)} \log \left| \frac{1+t}{1-t} \right| + c$.
Substituting $t = \tan x$ back,we get $I = \frac{1}{2} \log \left| \frac{1 + \tan x}{1 - \tan x} \right| + c$.

(C) Let $I = \int \csc^4 x \, dx$.
We can write $\csc^4 x$ as $\csc^2 x \cdot \csc^2 x$.
Using the identity $\csc^2 x = 1 + \cot^2 x$,we get:
$I = \int \csc^2 x (1 + \cot^2 x) \, dx$
$I = \int \csc^2 x \, dx + \int \cot^2 x \csc^2 x \, dx$
For the second integral,let $u = \cot x$,then $du = -\csc^2 x \, dx$,so $\csc^2 x \, dx = -du$.
$I = -\cot x + \int u^2 (-du)$
$I = -\cot x - \frac{u^3}{3} + c$
$I = -\cot x - \frac{\cot^3 x}{3} + c$.

(B) We use the identity $\cos^2 x = \frac{1 + \cos 2x}{2}$.
$\int x \cos^2 x \, dx = \int x \left( \frac{1 + \cos 2x}{2} \right) \, dx = \frac{1}{2} \int x \, dx + \frac{1}{2} \int x \cos 2x \, dx$.
Integrating the first part: $\frac{1}{2} \int x \, dx = \frac{x^2}{4}$.
For the second part,use integration by parts: $\int u \, dv = uv - \int v \, du$,where $u = x$ and $dv = \cos 2x \, dx$. Then $du = dx$ and $v = \frac{\sin 2x}{2}$.
$\frac{1}{2} \int x \cos 2x \, dx = \frac{1}{2} \left( \frac{x \sin 2x}{2} - \int \frac{\sin 2x}{2} \, dx \right) = \frac{x \sin 2x}{4} - \frac{1}{4} \int \sin 2x \, dx$.
$= \frac{x \sin 2x}{4} - \frac{1}{4} \left( -\frac{\cos 2x}{2} \right) = \frac{x \sin 2x}{4} + \frac{\cos 2x}{8}$.
Combining both parts: $\frac{x^2}{4} + \frac{x \sin 2x}{4} + \frac{\cos 2x}{8} + c$.

(D) We know that $\log_{10} x = \frac{\log_e x}{\log_e 10}$.
Thus,$\int \log_{10} x \, dx = \int \frac{\log_e x}{\log_e 10} \, dx = \frac{1}{\log_e 10} \int \log_e x \, dx$.
Using integration by parts,$\int \log_e x \, dx = x \log_e x - x + c$.
Therefore,$\int \log_{10} x \, dx = \frac{1}{\log_e 10} (x \log_e x - x) + c$.
$= \frac{x \log_e x}{\log_e 10} - \frac{x}{\log_e 10} + c$.
Since $\frac{\log_e x}{\log_e 10} = \log_{10} x$ and $\frac{1}{\log_e 10} = \log_{10} e$,we get:
$= x \log_{10} x - x \log_{10} e + c = x(\log_{10} x - \log_{10} e) + c$.

(C) We know that $\log_x e = \frac{1}{\log_e x}$.
Therefore,$\frac{1}{\log_x e} = \log_e x$.
The integral becomes $\int \log_e x \, dx$.
Using integration by parts,$\int u \cdot v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$,where $u = \log_e x$ and $v = 1$.
$\int \log_e x \cdot 1 \, dx = (\log_e x) \cdot x - \int \left( \frac{1}{x} \cdot x \right) \, dx$.
$= x \log_e x - \int 1 \, dx = x \log_e x - x + c$.
$= x(\log_e x - \log_e e) + c = x \log_e \left( \frac{x}{e} \right) + c$.

(D) To solve the integral $I = \int \frac{1}{\cos x(1 + \cos x)} \, dx$,we use partial fractions or algebraic manipulation.
We can write the integrand as: $\frac{1}{\cos x(1 + \cos x)} = \frac{(1 + \cos x) - \cos x}{\cos x(1 + \cos x)} = \frac{1}{\cos x} - \frac{1}{1 + \cos x}$.
Now,integrate each term separately:
$I = \int \sec x \, dx - \int \frac{1}{1 + \cos x} \, dx$.
Using the identity $1 + \cos x = 2 \cos^2 \frac{x}{2}$,we get:
$I = \log |\sec x + \tan x| - \int \frac{1}{2 \cos^2 \frac{x}{2}} \, dx$.
$I = \log |\sec x + \tan x| - \frac{1}{2} \int \sec^2 \frac{x}{2} \, dx$.
Integrating $\sec^2 \frac{x}{2}$ gives $2 \tan \frac{x}{2}$:
$I = \log |\sec x + \tan x| - \frac{1}{2} \cdot 2 \tan \frac{x}{2} + c$.
$I = \log |\sec x + \tan x| - \tan \frac{x}{2} + c$.

(B) We use the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$.
$\int \sin 5x \cos 3x \; dx = \frac{1}{2} \int (\sin(5x+3x) + \sin(5x-3x)) \; dx$
$= \frac{1}{2} \int (\sin 8x + \sin 2x) \; dx$
$= \frac{1}{2} \left( -\frac{\cos 8x}{8} - \frac{\cos 2x}{2} \right) + C$
$= -\frac{\cos 8x}{16} - \frac{\cos 2x}{4} + C$
Comparing this with the given expression $-\frac{\cos 8x}{16} + A$,we find that $A = -\frac{\cos 2x}{4} + C$.