Question

If the integral $\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{2k\sec \theta }}}=1-\dfrac{1}{\sqrt{2}}$ , (k>0) then find the value of k.

Answer 1

Hint: We start solving this problem first by considering the left-hand side of the given equation. Then we take out $\sqrt{2k}$ outside from the integral as it is a constant. We change $\tan \theta $ and $\sec \theta $ in terms of $\sin \theta $ and $\cos \theta $. Then we consider $\cos \theta $ as some other variable $t$ and we change the limits according to $t$. Then we solve the obtained integral. Hence, we get the value of $k$.

Complete step-by-step answer:
Let us consider the given equation $\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{2k\sec \theta }}}=1-\dfrac{1}{\sqrt{2}}.................\left( 1 \right)$
Now, we consider the left-hand side of the given equation, $\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{2k\sec \theta }}}$.
Let us take out $\sqrt{2k}$ outside from the integral as it is a constant.
$\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{2k\sec \theta }}}=\dfrac{1}{\sqrt{2k}}\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{\sec \theta }}}$
Let us consider the formula, $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\sec \theta =\dfrac{1}{\cos \theta }$.
By using the above formula, we change $\tan \theta $ and $\sec \theta $ in terms of $\sin \theta $ and $\cos \theta $, we get,
$\begin{align}
  & \int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{2k\sec \theta }}}=\dfrac{1}{\sqrt{2k}}\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{\sec \theta }}} \\
 & \\
 & \Rightarrow \int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{2k\sec \theta }}}=\dfrac{1}{\sqrt{2k}}\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\left( \dfrac{\sin \theta }{\cos \theta } \right)}{\sqrt{\dfrac{1}{\cos \theta }}}} \\
 & \\
 & \Rightarrow \int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{2k\sec \theta }}}=\dfrac{1}{\sqrt{2k}}\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\sin \theta }{\cos \theta }\times \sqrt{\cos \theta }} \\
 & \\
 & \Rightarrow \int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{2k\sec \theta }}}=\dfrac{1}{\sqrt{2k}}\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\sin \theta }{\sqrt{\cos \theta }}} \\
\end{align}$

Now, let us consider $\cos \theta =t$, by differentiating it on both the sides, we get,
$\begin{align}
  & -\sin \theta d\theta =dt \\
 & \Rightarrow \sin \theta d\theta =-dt \\
\end{align}$
Now, let us change the limits of the integral.
As the lower limit is 0 for $\theta $, the lower limit of the new integral is $t=\cos \theta =\cos \left( 0 \right)=1$ and
As the upper limit is $\dfrac{\pi }{3}$ for $\theta $, the upper limit for the new integral is $t=\cos \theta =\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$.
So, we get,
$\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{2k\sec \theta }}}=\dfrac{1}{\sqrt{2k}}\int\limits_{1}^{\dfrac{1}{2}}{\dfrac{-1}{\sqrt{t}}dt}$
Let us consider the formula, $\int{\dfrac{1}{\sqrt{x}}}=2\sqrt{x}$.
By using the above formula, we get,
$\begin{align}
  & \int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{2k\sec \theta }}}=\dfrac{1}{\sqrt{2k}}\left[ -2\sqrt{t} \right]_{1}^{\dfrac{1}{2}} \\
 & \\
 & \Rightarrow \int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{2k\sec \theta }}}=\dfrac{-1}{\sqrt{2k}}\left[ 2\sqrt{t} \right]_{1}^{\dfrac{1}{2}} \\
 & \\
 & \Rightarrow \int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{2k\sec \theta }}}=\dfrac{-1}{\sqrt{2k}}\left[ 2\sqrt{\dfrac{1}{2}}-2\sqrt{1} \right] \\
 & \\
 & \Rightarrow \int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{2k\sec \theta }}}=\dfrac{-1}{\sqrt{2k}}\left[ \dfrac{2}{\sqrt{2}}-2 \right] \\
 & \\
 & \Rightarrow \int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{2k\sec \theta }}}=\dfrac{-1}{\sqrt{2k}}\left[ \sqrt{2}-2 \right] \\
 & \\
 & \Rightarrow \int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{2k\sec \theta }}}=\dfrac{-1}{\sqrt{2}\sqrt{k}}\left[ \sqrt{2}-2 \right] \\
 & \\
 & \Rightarrow \int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{\tan \theta }{\sqrt{2k\sec \theta }}}=\dfrac{-1}{\sqrt{k}}\left[ 1-\sqrt{2} \right]....................\left( 2 \right) \\
\end{align}$

So, from equation (1) and equation (2), we get,
\[\begin{align}
  & \dfrac{-1}{\sqrt{k}}\left[ 1-\sqrt{2} \right]=1-\dfrac{1}{\sqrt{2}} \\
 & \\
 & \Rightarrow \dfrac{-1}{\sqrt{k}}\left[ 1-\sqrt{2} \right]=\dfrac{\sqrt{2}-1}{\sqrt{2}} \\
 & \\
 & \Rightarrow \dfrac{-1}{\sqrt{k}}=\dfrac{\sqrt{2}-1}{\sqrt{2}}\left( \dfrac{1}{1-\sqrt{2}} \right) \\
 & \\
 & \Rightarrow \dfrac{1}{\sqrt{k}}=\dfrac{1-\sqrt{2}}{\sqrt{2}}\left( \dfrac{1}{1-\sqrt{2}} \right) \\
 & \\
 & \Rightarrow \dfrac{1}{\sqrt{k}}=\dfrac{1}{\sqrt{2}} \\
 & \\
 & \Rightarrow k=2 \\
\end{align}\]

Therefore, the value of $k$ is 2.
Hence, the answer is 2.

Note: The possibility of making a mistake in this problem is one may make a mistake by not changing the limits while changing the variable. For example, in this problem, while changing the variable from $\theta $ to $t$, we change the lower limit from 0 to 1 and upper limit from $\dfrac{\pi }{3}$ to $\dfrac{1}{2}$. Otherwise, we get the wrong answer.