Question

The value of the integral $\int\limits_{-\dfrac{\pi }{3}}^{\dfrac{\pi }{3}}{\dfrac{x\sin x}{{{\cos }^{2}}x}dx}$ is
(a) $\left( \dfrac{\pi }{3}-\log \tan \dfrac{3\pi }{2} \right)$,
(b) $2\left( \dfrac{2\pi }{3}-\log \tan \dfrac{5\pi }{12} \right)$,
(c) $3\left( \dfrac{\pi }{2}-\log \tan \dfrac{\pi }{12} \right)$,
(d) none of these.

Answer 1

Hint: To solve the question given above, we will change the limit form \[\left( -\dfrac{\pi }{3}\ to\text{ }\dfrac{\pi }{3} \right)to\left( 0\text{ }to\ \dfrac{2\pi }{3} \right)\] with the help of the fact that the term after integration is function. Then we will change the term $\dfrac{\sin x}{{{\cos }^{2}}x}$ such that we can apply by parts to the integral and after applying by parts and solving the indefinite part, we will apply limit.

Complete step by step answer:
To start with, we will find whether the function is odd or even. For this, we will put (-x) in place of x. Let $f\left( x \right)=\dfrac{x\sin x}{{{\cos }^{2}}x}.$ Thus, $f\left( -x \right)$ will be:
$\begin{align}
  & \Rightarrow f\left( -x \right)=\dfrac{\left( -x \right)\sin \left( -x \right)}{{{\cos }^{2}}x} \\
 & \Rightarrow f\left( -x \right)=\dfrac{-x\sin \left( -x \right)}{{{\cos }^{2}}x} \\
\end{align}$
Now, we know that $\sin \left( -x \right)=-\sin x.$ Thus, we will get:
$\begin{align}
  & \Rightarrow f\left( -x \right)=\dfrac{-x\left( -\sin x \right)}{{{\cos }^{2}}x} \\
 & \Rightarrow f\left( -x \right)=\dfrac{x\sin x}{{{\cos }^{2}}} \\
 & \Rightarrow f\left( -x \right)=f\left( x \right) \\
\end{align}$
Hence it is an even function.
Now, we know that, if a function g (x) is an even function and the integral given is of the form:$\int\limits_{-a}^{a}{g\left( x \right)dx}$ then, we will have:
$\int\limits_{-a}^{a}{g\left( x \right)dx}=2\int\limits_{-a}^{a}{g\left( x \right)dx}$.
Let I be the value of the integral given in the question. Thus, we have:
$\begin{align}
  & I=\int\limits_{-\dfrac{\pi }{3}}^{\dfrac{\pi }{3}}{\dfrac{x\sin x}{{{\cos }^{2}}x}dx} \\
 & \Rightarrow I=2\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{x\sin x}{{{\cos }^{2}}x}dx}.........\left( 1 \right) \\
\end{align}$
Now, we know that , $f\left( x \right)=\dfrac{x\sin x}{{{\cos }^{2}}x}.........\left( 2 \right)$.
We can also write this as:
$\begin{align}
  & \Rightarrow f\left( x \right)=\dfrac{x\sin x}{\left( \cos x \right)\left( \cos x \right)} \\
 & \Rightarrow f\left( x \right)=\left( \dfrac{x}{\cos x} \right)\left( \dfrac{\sin x}{\cos x} \right) \\
\end{align}$
Now, we know that: $\dfrac{1}{\cos x}=\sec x\text{ }and\text{ }\dfrac{\sin x}{\cos x}=\tan x$, we will put these values in the above equation thus, we get:
$\begin{align}
  & \Rightarrow f\left( x \right)=\left( x\sec x \right)\left( \tan x \right) \\
 & \Rightarrow f\left( x \right)=x\sec xtanx.........\left( 3 \right) \\
\end{align}$
From (2) and (3), we have:
$\dfrac{x\sin x}{{{\cos }^{2}}x}=x\sec x\tan x$.
Now, we will put the value of $\dfrac{x\sin x}{{{\cos }^{2}}x}$ in equation (1). Thus, we will get:
\[\Rightarrow I=2\int\limits_{0}^{\dfrac{\pi }{3}}{x\sec x\text{ }\tan x\ dx}\].
Now to solve this integration, we will use the method of by-parts. According to by-parts method, we have: $\int{\left( \operatorname{uv} \right)dx}=v\int{\operatorname{u}dx}-\int{\left( \dfrac{dv}{dx}\times \int{\operatorname{u}dx} \right)dx}$. We also know while performing an integration for definite integral, we substitute limits after normal integration to the integrand i.e., $\int\limits_{a}^{b}{{{f}^{'}}\left( x \right)dx}=\left[ f\left( x \right) \right]_{a}^{b}$.
In our case, we will take $u=x\ and\ v=\sec x\ \tan x.$ Thus, we will get:
$\begin{align}
  & \Rightarrow I=2\int\limits_{0}^{\dfrac{\pi }{3}}{\left( x \right)\left( \sec x\tan x \right)dx} \\
 & \Rightarrow I=2\left[ x\int{\sec x\tan xdx-\int{\dfrac{dx}{dx}\int{\sec x\tan x{{\left( dx \right)}^{2}}}}} \right]_{0}^{\dfrac{\pi }{3}} \\
 & \Rightarrow I=2\left[ x\int{\sec x\tan xdx-\int{1\int{\sec x\tan x{{\left( dx \right)}^{2}}}}} \right]_{0}^{\dfrac{\pi }{3}}.........\left( 4 \right) \\
\end{align}$
Now, we know that differentiation of $\sec x$ is given by:
$\begin{align}
  & \dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x \\
 & \Rightarrow \sec x=\int{\sec x\tan xdx} \\
 & \Rightarrow \int{\sec x\tan xdx}=\sec x+c.........\left( 5 \right) \\
\end{align}$
Now, we will put the value of $\int{\sec x\tan x}$ from (5) to (4). Thus, we will get:
$\Rightarrow I=2\left[ x\left( \sec x \right)-\int{\sec xdx} \right]_{0}^{\dfrac{\pi }{3}}..........\left( 6 \right)$.
Now, the integration of $\sec \theta $ is given by:
$\int{\sec \theta d\theta }=\log \left( \sec \theta +\tan \theta \right)+c$.
Thus, using the above formula in (6), we will get:
$\Rightarrow I=2\left[ x\sec x-\log \left( \sec x+\tan x \right) \right]_{0}^{\dfrac{\pi }{3}}$.
Now, we have to find the indefinite integral so we will apply the limits on this indefinite integral we calculated. The limit of any integral can be applied by:
 $\int{A\left( x \right)}=\left[ B\left( x \right) \right]_{a}^{b}=B\left( b \right)-B\left( a \right)$.
Thus, we will have:
$\begin{align}
  & \Rightarrow I=2\left[ \dfrac{\pi }{3}\sec \left( \dfrac{\pi }{3} \right)-\log \left( \sec \dfrac{\pi }{3}+\tan \dfrac{\pi }{3} \right)-\left( 0\sec 0-\log \left( \sec 0+\tan 0 \right) \right) \right] \\
 & \Rightarrow I=2\left[ \dfrac{\pi }{3}\sec \left( \dfrac{\pi }{3} \right)-\log \left( \sec \dfrac{\pi }{3}+\tan \dfrac{\pi }{3} \right)-0\sec 0+\log \left( \sec 0+\tan 0 \right) \right] \\
\end{align}$
Now, we will put the values of $\sec \dfrac{\pi }{3},\tan \dfrac{\pi }{3},\sec 0,\tan 0$ is the above integral:
$\begin{align}
  & \sec \dfrac{\pi }{3}=2 \\
 & \tan \dfrac{\pi }{3}=\sqrt{3} \\
 & \sec 0=1 \\
 & \tan 0=0 \\
\end{align}$
So, we have:
$\begin{align}
  & \Rightarrow I=2\left[ \dfrac{\pi \left( 2 \right)}{3}-\log \left( 2+\sqrt{3} \right)-0\left( 0 \right)+\log \left( 1+0 \right) \right] \\
 & \Rightarrow I=2\left[ \dfrac{2\pi }{3}-\log \left( 2+\sqrt{3} \right)-0+0 \right] \\
 & \Rightarrow I=2\left[ \dfrac{2\pi }{3}-\log \left( 2+\sqrt{3} \right) \right]...............\left( 7 \right) \\
\end{align}$
Now, we know that: $\tan \dfrac{5\pi }{12}=2+\sqrt{3}.........\left( 8 \right).$
 Thus, we will put the value of $\left( 2+\sqrt{3} \right)$ in equation (7) from (8):
$\Rightarrow I=2\left[ \dfrac{2\pi }{3}-\log \tan \dfrac{5\pi }{12} \right]$.

So, the correct answer is “Option B”.

Note: The term given in the question after the integration sign is an even function. If the function had been an odd function then we could not have used the identity:
\[\int\limits_{-a}^{a}{g\left( x \right)dx}=2\int\limits_{0}^{a}{g\left( x \right)dx}\].
If the function \[g\left( x \right)\] is odd function, then we have:
\[\int\limits_{-a}^{a}{g\left( x \right)dx}=0\].
We can solve the question without using the concept of odd and even functions. We can simply solve the question and put the limit from $\left( -\dfrac{\pi }{3}to\dfrac{\pi }{3} \right).$