Question

$\int \frac{\cos 4x+1}{\cot x-\tan x}dx=k\cos 4x+C$, then

• A. $k=-\frac{1}{2}$
• B. $k=-\frac{1}{8}$
• C. $k=-\frac{1}{4}$
• D. None of these

Answer 1

The correct option is B

$k=-\frac{1}{8}$

Explanation for Correct answer:

Solving the equation to find the value of $k$:

Given,

$\int \frac{\cos 4x+1}{\cot x-\tan x}dx=k\cos 4x+C\to \left(i\right)$

$$\begin{gathered} \int \frac{\cos 4x+1}{\cot x-\tan x}dx=\int \frac{2{\cos }^{2}2x}{{\displaystyle \frac{c\mathrm{os}x}{\sin x}}-{\displaystyle \frac{\sin x}{\cos x}}}dx\left[\because \cos 2x=2{\cos }^{2}x-1\right] \\ =\int \frac{2{\cos }^{2}2x}{{\displaystyle \frac{{\cos }^{2}x-{\sin }^{2}x}{\sin x\cos x}}}dx \\ =\int \frac{2{\cos }^{2}2x}{{\displaystyle \frac{2\cos 2x}{\sin 2x}}}dx\left[\because {\cos }^{2}x-{\sin }^{2}x=\cos 2x;\sin 2x=2\sin x\cos x\right] \\ =\int \cos 2x\sin 2xdx \\ =\int \frac{\sin 4x}{2}dx\left[\because \sin 2x=2\sin x\cos x\right] \\ =-\frac{1}{2}\frac{\cos 4x}{4}+C \\ =-\frac{1}{8}\cos 4x+C\to \left(ii\right) \end{gathered}$$

Equating the equations $\left(i\right)$ and $\left(ii\right)$

$$\begin{gathered} k\cos 4x+C=-\frac{1}{8}\cos 4x+C \\ \Rightarrow k=-\frac{1}{8} \end{gathered}$$

Hence, option (B) is the correct answer.