Question

In the integral $\int{\dfrac{\cos 8x+1}{\cot 2x-\tan 2x}dx=A\cos 8x+k}$, where $k$ is an arbitrary constant, then the value of $k$ is equal to
A. $-\dfrac{1}{16}$
B. $\dfrac{1}{16}$
C. $\dfrac{1}{8}$
D. $-\dfrac{1}{8}$

Answer 1

Hint: We will first substitute the value of trigonometric ratios $\cot 2x=\dfrac{\cos 2x}{\sin 2x}$, $\tan 2x=\dfrac{\sin 2x}{\cos 2x}$ in the denominator of the given integral. After substituting the values we will simplify the integral and then we will use the formulas $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$ and $\sin 2x=2\sin x\cos x$ in the integral and then we will integrated the simplified equation and compare the result with the given equation to find the value of $A$.

Complete step by step answer:
Given that, $\int{\dfrac{\cos 8x+1}{\cot 2x-\tan 2x}dx=A\cos 8x+k}$
Taking L.H.S into consideration, then
$L.H.S=\int{\dfrac{\cos 8x+1}{\cot 2x-\tan 2x}dx}$
We know that $\cot 2x=\dfrac{\cos 2x}{\sin 2x}$, $\tan 2x=\dfrac{\sin 2x}{\cos 2x}$ substituting these values in the above equation, then we will get
$L.H.S=\int{\dfrac{\cos 8x+1}{\dfrac{\cos 2x}{\sin 2x}-\dfrac{\sin 2x}{\cos 2x}}dx}$
Simplifying the denominator by taking LCM of $\sin 2x$ and $\cos 2x$, then we will get
$L.H.S=\int{\dfrac{\cos 8x+1}{\dfrac{\cos 2x.\cos 2x-\sin 2x.\sin 2x}{\sin 2x.\cos 2x}}dx}$
We know that $a.a={{a}^{2}}$ and writing the $\cos 8x$ as $\cos 2\left( 4x \right)$, then we will get
$L.H.S=\int{\dfrac{\cos 2\left( 4x \right)+1}{\dfrac{{{\cos }^{2}}2x-{{\sin }^{2}}2x}{\sin 2x.\cos 2x}}dx}$
We have the formula $\cos 2x=2{{\cos }^{2}}x-1={{\cos }^{2}}x-{{\sin }^{2}}x$, now the above equation modified as
$\begin{align}
  & L.H.S=\int{\dfrac{2{{\cos }^{2}}4x-1+1}{\dfrac{\cos 2\left( 2x \right)}{\sin 2x.\cos 2x}}dx} \\
 & \Rightarrow L.H.S=\int{\dfrac{2{{\cos }^{2}}4x\left( \sin 2x.\cos 2x \right)}{\cos 4x}dx} \\
 & \Rightarrow L.H.S=\int{\cos 4x\left( 2\sin 2x.\cos 2x \right)dx} \\
\end{align}$
We have the value of $2\sin x.\cos x=\sin 2x$, then we will get
$\begin{align}
  & \Rightarrow L.H.S=\int{\cos 4x\left( \sin 2\left( 2x \right) \right)}dx \\
 & \Rightarrow L.H.S=\int{\cos 4x.\sin 4x.dx} \\
\end{align}$
Multiplying and dividing the above equation with $2$, then we will have
$\Rightarrow L.H.S=\dfrac{1}{2}\int{2.\sin 4x.\cos 4x}dx$
Again, using the formula $2\sin x.\cos x=\sin 2x$ in the above equation, then we will get
$\begin{align}
  & \Rightarrow L.H.S=\dfrac{1}{2}\int{\sin 2\left( 4x \right)}dx \\
 & \Rightarrow L.H.S=\dfrac{1}{2}\int{\sin 8x.dx} \\
\end{align}$
We know that $\int{\sin ax.dx}=-\dfrac{\cos ax}{a}+C$, then we will get
$\begin{align}
  & L.H.S=\dfrac{1}{2}\left[ -\dfrac{\cos 8x}{8}+C \right] \\
 & \Rightarrow L.H.S=-\dfrac{1}{16}\cos 8x+k \\
\end{align}$
Now comparing the LHS and RHS of the given equation.
$-\dfrac{1}{16}\cos 8x+k=A\cos 8x+k$
Equating on both sides we will get the value of $A$ as $A=-\dfrac{1}{16}$.

So, the correct answer is “Option A”.

Note: When we are dealing with the integrals including the trigonometric ratios, we need to know about some basic trigonometric formulas, like trigonometric identities, half angle formulas etc. Some of the formulas are listed below
$\begin{align}
  & {{\sin }^{2}}x-{{\cos }^{2}}x=1 \\
 & {{\sec }^{2}}x-{{\tan }^{2}}x=1 \\
 & {{\csc }^{2}}x-{{\cot }^{2}}x=1 \\
 & \tan x=\dfrac{\sin x}{\cos x} \\
 & \sec x=\dfrac{1}{\cos x} \\
 & \csc x=\dfrac{1}{\sin x} \\
 & \cot x=\dfrac{1}{\tan x}=\dfrac{\cos x}{\sin x} \\
 & \sin 2x=2\sin x\cos x \\
 & \cos 2x=2{{\cos }^{2}}x-1=1-2{{\sin }^{2}}x={{\cos }^{2}}x-{{\sin }^{2}}x \\
 & \tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x} \\
\end{align}$