Question

\[\int {\frac{{1 + {{\cot }^2}x}}{{1 + \cot x}}dx = } \]
A. \[ - \log\lvert1+cot x\lvert+c\]
B. \[\log\lvert1+tan x\lvert+c\]
C. \[\log(1 + cot x) + c\]
D. \[\log(cot x) + c\]

Answer 1

Hint: Hint: We have to solve the given integration. First we will use the formula \[1 + {\cot ^2}x = {cosec^2}x\] . Next we will go for substitution method to get our desired solution.

Formula used: Trigonometric formula: \[\begin{array}{l}\cos e{c^2}x - {\cot ^2}x = 1\\or,\cos e{c^2}x = 1 + {\cot ^2}x\end{array}\]
Integration formula: \[\int {\frac{{dx}}{x} = \log \left| x \right|} \]

Complete step-by-step solution:
We have to solve the given integration. First we will use the formula \[1 + {\cot ^2}x = \cos e{c^2}x\] . Next we will go for substitution method to get our desired solution.
\[\begin{array}{l}\int {\frac{{1 + {{\cot }^2}x}}{{1 + \cot x}}dx} \\ = \int {\frac{{\cos e{c^2}x}}{{1 + \cot x}}dx} \end{array}\]
[Substitute \[1 + \cot x = z\] .Taking differential in both sides we have \[ - \cos e{c^2}xdx = dz\] ]
\[\begin{array}{l} = - \int {\frac{{dz}}{z}} \\ = - \log \left| z \right| + C\\ = - \log \left| {1 + \cot x} \right| + C\end{array}\] \[\] [\[\int {\frac{{dx}}{x} = \log \left| x \right|} \] ]
 where C is integration constant.

The option (a) is correct.

Note: Students often make the mistake of not transferring the new variable to the old one. It is very important to change to an old variable. A new variable has been taken for solving our problems. This has no role in the given problem. So, you must keep the answer in the old variable. Students also forget to write integration constant after integration. It should be written. So, students! Keep remembering all of these.