Question

If $\int {\cos ec2xdx = f\left( {g\left( x \right)} \right) + C} $, then
A) range $g\left( x \right) = \left( { - \infty ,\infty } \right)$
B) dom $g\left( x \right) = \left( { - \infty ,\infty } \right) - \left\{ 0 \right\}$
C) $g'\left( x \right) = {\sec ^2}x$
D) $f'\left( x \right) = \dfrac{1}{x}$ for all $x \in \left( {0,\infty } \right)$

Answer 1

Hint:
Our given function can be written as \[\int {\dfrac{{dx}}{{\sin 2x}}} \] and by using the identity $\sin 2x = 2\sin x\cos x$ we get an integral and we can use our substitution method by taking $t = \sin x$ and simplifying further once again we use substitution method by taking $y = \left( {1 - {t^2}} \right)$ and further simplifying we get $f(x) = \ln x$ and $g(x) = \tan x$ using this we get our required conditions.

Complete step by step solution:
We are given a function \[\int {\cos ec2xdx} \]
We know that $\cos ec x$ can be written as $\dfrac{1}{{\sin x}}$
Using this we get
 \[ \Rightarrow \int {\dfrac{{dx}}{{\sin 2x}}} \]
Then let's use the identity $\sin 2x = 2\sin x\cos x$
\[ \Rightarrow \int {\dfrac{{dx}}{{2\sin x\cos x}}} \]
Now lets use substitution method to solve the following integral
Let $t = \sin x$
Differentiating we get $dt = \cos xdx$
\[
   \Rightarrow \int {\dfrac{{dt}}{{2t\cos x\cos x}}} \\
   \Rightarrow \int {\dfrac{{dt}}{{2t{{\cos }^2}x}}} \\
 \]
Now since $t = \sin x$
$
   \Rightarrow {t^2} = {\sin ^2}x \\
   \Rightarrow {t^2} = 1 - {\cos ^2}x \\
   \Rightarrow {\cos ^2}x = 1 - {t^2} \\
 $
 Using this we get
\[ \Rightarrow \int {\dfrac{{dt}}{{2t\left( {1 - {t^2}} \right)}}} \]
Now let's add and subtract in the numerator
\[
   \Rightarrow \int {\dfrac{{1 + {t^2} - {t^2}}}{{2t\left( {1 - {t^2}} \right)}}dt} \\
   \Rightarrow \int {\dfrac{{1 - {t^2}}}{{2t\left( {1 - {t^2}} \right)}} + } \dfrac{{{t^2}}}{{2t\left( {1 - {t^2}} \right)}}dt \\
   \Rightarrow \int {\dfrac{1}{{2t}} + } \dfrac{t}{{2\left( {1 - {t^2}} \right)}}dt \\
 \]
Now let's multiply and divide 2 in the second part
\[ \Rightarrow \int {\dfrac{1}{{2t}}dt} + \dfrac{1}{4}\int {\dfrac{{2t}}{{\left( {1 - {t^2}} \right)}}dt} \]
Integrating the first part we get
 \[ \Rightarrow \dfrac{1}{2}\ln t + \dfrac{1}{4}\int {\dfrac{{2t}}{{\left( {1 - {t^2}} \right)}}dt} \]
Now once again let's use substitution method
Let $y = \left( {1 - {t^2}} \right)$
Differentiating we get $dy = - 2tdt$
Now we get
 \[ \Rightarrow \dfrac{1}{2}\ln t + \dfrac{1}{4}\int {\dfrac{{ - dy}}{y}} \]
Integrating this we get
\[ \Rightarrow \dfrac{1}{2}\ln t + \dfrac{{ - 1}}{4}\ln y + C\]
Replacing y by $\left( {1 - {t^2}} \right)$
\[
   \Rightarrow \dfrac{1}{2}\ln t - \dfrac{1}{4}\ln \left( {1 - {t^2}} \right) + C \\
   \Rightarrow \dfrac{{2\ln t - \ln \left( {1 - {t^2}} \right)}}{4} + C \\
   \Rightarrow \dfrac{{\ln {t^2} - \ln \left( {1 - {t^2}} \right)}}{4} + C \\
   \Rightarrow \dfrac{{\ln \left( {\dfrac{{{t^2}}}{{\left( {1 - {t^2}} \right)}}} \right)}}{4} + C \\
 \]
Now using $t = \sin x$
\[
   \Rightarrow \dfrac{{\ln \left( {\dfrac{{{{\sin }^2}x}}{{\left( {1 - {{\sin }^2}x} \right)}}} \right)}}{4} + C \\
   \Rightarrow \dfrac{{\ln \left( {\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}}{4} + C \\
   \Rightarrow \dfrac{{\ln \left( {{{\tan }^2}x} \right)}}{4} + C \\
   \Rightarrow \dfrac{{2\ln \left( {\tan x} \right)}}{4} + C \\
   \Rightarrow \dfrac{{\ln \left( {\tan x} \right)}}{2} + C \\
 \]
From this obtain that $f(x) = \ln x$ and $g(x) = \tan x$
Differentiating them we get $f'(x) = \dfrac{1}{x}$ and $g'(x) = {\sec ^2}x$
Now with our g(x) we can see the range and domain of tanx is $\left( { - \infty ,\infty } \right)$
From this we get that the options a , c and d are correct.

Note:
The domain of a function f(x) is the set of all values for which the function is defined, and the range of the function is the set of all values that f takes. (In grammar school, you probably called the domain the replacement set and the range the solution set.