Use the substitution \(u= x\sin x +\cos x\) to find
\[
\int \frac{x }{x\tan x +1 } \, \d x
\,.
\]
Find by means of a similar substitution, or otherwise,
\[
\int \frac{x }{x\cot x -1 } \, \d x
\,.
\]
Use a substitution to find
\[
\int \frac{x\sec^2 x \, \tan x}{x\sec^2 x -\tan x} \,\d x
\,
\]
and
\[
\int \frac{x\sin x \cos x}{(x-\sin x \cos x)^2} \, \d x
\,.
\]
Solution:
\(\,\)
\begin{align*}
&& I &= \int \frac{x}{x \tan x + 1} \d x \\
&&&= \int \frac{x \cos x}{x \sin x + \cos x} \d x \\
u = x \sin x + \cos x , \d u = x \cos x \d x: &&&= \int \frac{\d u}{u} \\
&&&= \ln u + C \\
&&&= \ln (x \sin x + \cos x) + C \\
\\
&& J &= \int \frac{x}{x \cot x - 1} \d x \\
&&&= \int \frac{x \sin x }{x \cos x - \sin x} \d x \\
u = x \cos x - \sin x, \d u = x \sin x \d x: &&&= \int \frac{1}{u} \d u \\
&&&= \ln u + K \\
&&&= \ln (x \cos x -\sin x) + K
\end{align*}
\(\,\)
\begin{align*}
&& I &= \int \frac{x \sec^2 x \tan x}{x \sec^2 x - \tan x} \d x \\
u = x\sec^2 x-\tan x, \d u = 2x \sec^2 x \tan x&&&= \frac12 \int \frac{1}{u} \d u \\
&&&= \frac12 \ln (x \sec^2 x - \tan x) + C \\
\\
&& J &= \int \frac{x \sin x \cos x}{(x - \sin x \cos x)^2} \d x \\
u = x \sec^2 x -\tan x, \d u=2x \frac{\sin x}{\cos^3 x} &&&= \int \frac{x \sin x \cos x}{\cos^4x(x\sec^2 x -\tan x)^2} \d x \\
&&&= \frac12 \int \frac{1}{u^2} \d u \\
&&&= -\frac12u^{-1} + K \\
&&&= \frac{1}{2(\tan x - x \sec^2 x)} + K
\end{align*}