Question

The value of \[\int\limits_{\dfrac{1}{e}}^{\tan x}{\dfrac{tdt}{1+{{t}^{2}}}}+\int\limits_{\infty }^{\cot x}{\dfrac{dt}{t(1+{{t}^{2}})}}\] where \[x\in \left( \dfrac{\pi }{6},\dfrac{\pi }{3} \right)\] equals
A. 0
B. 2
C. 1
D. -1

Answer 1

Hint: Assume the given expression as \[I={{I}_{1}}+{{I}_{2}}\] where, \[{{I}_{1}}=\int\limits_{\dfrac{1}{e}}^{\tan x}{\dfrac{tdt}{1+{{t}^{2}}}}\] and \[{{I}_{2}}=\int\limits_{\infty }^{\cot x}{\dfrac{dt}{t(1+{{t}^{2}})}}\] . Now, solve \[{{I}_{1}}\] and \[{{I}_{2}}\] separately, Transform \[{{I}_{1}}\] by assuming \[P=1+{{t}^{2}}\] and then get the upper and lower limits of the definite integral. Similarly, transform \[{{I}_{2}}\] by assuming \[t=\tan \beta \] and then get the upper and lower limits of the definite integral. Now, using Pythagoras theorem change the upper limit of \[{{I}_{2}}\] to the inverse of sine. Now, put the values that we have got after solving \[{{I}_{1}}\] and \[{{I}_{2}}\] in the equation
\[I={{I}_{1}}+{{I}_{2}}\] and then solve it further.

Complete step-by-step answer:
First of all, let us assume
\[{{I}_{1}}=\int\limits_{\dfrac{1}{e}}^{\tan x}{\dfrac{tdt}{1+{{t}^{2}}}}\] …………………………(1)
\[{{I}_{2}}=\int\limits_{\infty }^{\cot x}{\dfrac{dt}{t(1+{{t}^{2}})}}\] …………………….(2)
\[I=\int\limits_{\dfrac{1}{e}}^{\tan x}{\dfrac{tdt}{1+{{t}^{2}}}}+\int\limits_{\infty }^{\cot x}{\dfrac{dt}{t(1+{{t}^{2}})}}\] …………………………(3)
Transforming equation (3), we get
\[I={{I}_{1}}+{{I}_{2}}\] …………………..(4)
Now, solving equation (1),
\[{{I}_{1}}=\int\limits_{\dfrac{1}{e}}^{\tan x}{\dfrac{tdt}{1+{{t}^{2}}}}\]
Let us assume,
\[P=1+{{t}^{2}}\] ………………..(5)
On differentiating equation (5), we get
\[\dfrac{dP}{dt}=2t\]
\[\Rightarrow \dfrac{1}{2}dP=2tdt\] …………………….(6)
We have upper limits and lower limits for t and here, it is required to find upper and lower limits of P.
Lower limit of t is \[\dfrac{1}{e}\] .
From equation (5), we have, \[P=1+{{t}^{2}}\] .
If \[t=\dfrac{1}{e}\] , then P will be
\[P=1+\dfrac{1}{{{e}^{2}}}\]
So, lower limit of P is \[1+\dfrac{1}{{{e}^{2}}}\] ………………..(7)
Upper limit of t is \[\tan x\] .
From equation (5), we have, \[P=1+{{t}^{2}}\] .
We know the identity, \[{{\sec }^{2}}x-{{\tan }^{2}}=1\] .
\[\begin{align}
  & {{\sec }^{2}}x-{{\tan }^{2}}x=1 \\
 & \Rightarrow {{\sec }^{2}}x=1+{{\tan }^{2}}x \\
\end{align}\]
Now using the above identity and we have the upper limit \[t=\tan x\] , then P will be
\[P=1+{{\tan }^{2}}x={{\sec }^{2}}x\]
So, upper limit of P is \[{{\sec }^{2}}x\] ……………(8)
\[{{I}_{1}}=\int\limits_{\dfrac{1}{e}}^{\tan x}{\dfrac{tdt}{1+{{t}^{2}}}}\]
Transforming the above equation using equation (5), equation (6), equation (7), and equation (8), we get
\[{{I}_{1}}=\dfrac{1}{2}\int\limits_{\dfrac{1}{e}}^{\tan x}{\dfrac{2tdt}{1+{{t}^{2}}}}\]
\[\begin{align}
  & {{I}_{1}}=\dfrac{1}{2}\int\limits_{1+\dfrac{1}{{{e}^{2}}}}^{{{\sec }^{2}}x}{\dfrac{1}{P}}dP \\
 & \Rightarrow {{I}_{1}}=\dfrac{1}{2}\left[ \ln P \right]_{1+\dfrac{1}{{{e}^{2}}}}^{{{\sec }^{2}}x} \\
 & \Rightarrow {{I}_{1}}=\dfrac{1}{2}\left[ \ln {{\sec }^{2}}x-\ln (1+\dfrac{1}{{{e}^{2}}}) \right] \\
 & \Rightarrow {{I}_{1}}=\dfrac{1}{2}\ln \dfrac{{{\sec }^{2}}x.{{e}^{2}}}{(1+{{e}^{2}})} \\
 & \Rightarrow {{I}_{1}}=\ln \dfrac{\sec x.e}{\sqrt{(1+{{e}^{2}})}} \\
\end{align}\]
\[\Rightarrow {{I}_{1}}=\ln \dfrac{\sec x}{\sqrt{\dfrac{1}{{{e}^{2}}}+1}}\] ………………………….(9)
Now, solving equation (2),
\[{{I}_{2}}=\int\limits_{\infty }^{\cot x}{\dfrac{dt}{t(1+{{t}^{2}})}}\]
Let us assume,
\[t=\tan \beta \] …………………(10)
On differentiating equation (10), we get
\[\dfrac{dt}{d\beta }={{\sec }^{2}}\beta \]
\[\Rightarrow dt={{\sec }^{2}}\beta d\beta \] ………………..(11)
We have upper limits and lower limits for t and here, it is required to find upper and lower limits of
\[\beta \] .
Upper limit of t is \[\cot x\] .
From equation (10), we have, \[t=\tan \beta \] .
To get upper limit of \[\beta \], we have \[\tan \beta =\cot x\] , then \[\beta \] will be
\[t=\tan \beta =\cot x\]
\[\Rightarrow \beta ={{\tan }^{-1}}(cotx)\] ……………………….(12)
In equation (12) we have inverse of tan. We need to convert it into inverse of sine.
\[\tan \beta =\cot x\]
We know that, \[\tan \beta =\dfrac{Perpendicular}{Base}\] and \[\sin \beta =\dfrac{Perpendicular}{Hypotenuse}\] .
\[\tan \beta =\dfrac{Perpendicular}{Base}=\dfrac{\cot x}{1}\]
Using, Pythagoras theorem,
\[\begin{align}
  & {{(Hypotenuse)}^{2}}={{\left( Perpendicular \right)}^{2}}+{{\left( Base \right)}^{2}} \\
 & \Rightarrow {{(Hypotenuse)}^{2}}={{\left( \cot x \right)}^{2}}+{{1}^{2}} \\
 & \Rightarrow {{(Hypotenuse)}^{2}}={{\cot }^{2}}x+1 \\
 & \Rightarrow Hypotenuse=\sqrt{{{\cot }^{2}}x+1} \\
\end{align}\]
We know the identity, \[{{\operatorname{cosec}}^{2}}x-{{\cot }^{2}}x=1\Rightarrow {{\operatorname{cosec}}^{2}}x=1+{{\cot }^{2}}x\] .
Using this identity and formula of \[\sin \beta \] , we get
\[\sin \beta =\dfrac{Perpendicular}{Hypotenuse}=\dfrac{\cot x}{1+{{\cot }^{2}}x}=\dfrac{\cot x}{\operatorname{cosec}x}\] ………………..(13)
We know that, \[\cot x=\dfrac{\cos x}{\sin x}\] and \[\operatorname{cosec}x=\dfrac{1}{\sin x}\] . Now, using this in equation (13), we get
\[\sin \beta =\dfrac{\cot x}{\operatorname{cosec}x}=\dfrac{\dfrac{\cos x}{\sin x}}{\dfrac{1}{\sin x}}\]
\[\begin{align}
  & \sin \beta =\cos x \\
 & \Rightarrow \beta ={{\sin }^{-1}}\left( \cos x \right) \\
\end{align}\]
So, our upper limit is \[\beta ={{\sin }^{-1}}\left( \cos x \right)\] ………………(14)
Lower limit of \[\beta \] is \[\infty \] .
From equation (10), we have, \[t=\tan \beta \] .
To get upper limit of \[\beta \], we have \[\tan \beta =\infty \] , then \[\beta \] will be
\[t=\tan \beta =\infty \] ……………(15)
We know the that, \[\tan \dfrac{\pi }{2}=\infty \] and the property \[{{\tan }^{-1}}\left( \tan \beta \right)=\beta \] .
Using this in equation (15), we get
\[\begin{align}
  & t=\tan \beta =\infty \\
 & \Rightarrow \beta ={{\tan }^{-1}}(tan\dfrac{\pi }{2})=\dfrac{\pi }{2} \\
\end{align}\]
So, our lower limit is \[\beta =\dfrac{\pi }{2}\] ………………(16)
On transforming equation (2), using equation (10), equation (11), equation (14), and equation (16), we get,
\[{{I}_{2}}=\int\limits_{\infty }^{\cot x}{\dfrac{dt}{t(1+{{t}^{2}})}}\]
We know the identity that, \[{{\sec }^{2}}x-{{\tan }^{2}}x=1\Rightarrow {{\sec }^{2}}x=1+{{\tan }^{2}}x\] . Using this identity in the above equation, we get
\[{{I}_{2}}=\int\limits_{\dfrac{\pi }{2}}^{{{\sin }^{-1}}(cosx)}{\dfrac{{{\sec }^{2}}\beta d\beta }{\tan \beta (1+ta{{n}^{2}}\beta )}}\]
\[\begin{align}
  & =\int\limits_{\dfrac{\pi }{2}}^{{{\sin }^{-1}}(cosx)}{\dfrac{1}{\tan \beta .{{\sec }^{2}}\beta }}{{\sec }^{2}}\beta d\beta \\
 & =\int\limits_{\dfrac{\pi }{2}}^{{{\sin }^{-1}}(cosx)}{\dfrac{1}{\tan \beta }d\beta } \\
 & =\int\limits_{\dfrac{\pi }{2}}^{{{\sin }^{-1}}(cosx)}{\cot \beta d\beta } \\
\end{align}\]
We know the formula that, \[\int{\cot xdx=\ln (sinx)}\] . Using this formula in the above equation, we get
\[\begin{align}
  & =\left[ \ln \left( \sin \beta \right) \right]_{\dfrac{\pi }{2}}^{{{\sin }^{-1}}(cosx)} \\
 & =\left[ \ln \left\{ \sin \left( {{\sin }^{-1}}(cosx) \right) \right\}-\ln \sin \dfrac{\pi }{2} \right] \\
 & =\ln (cosx)-\ln 1 \\
 & =\ln (cosx) \\
\end{align}\]
\[{{I}_{2}}=\ln (cosx)\] …………………..(17)
From equation (9), we have \[{{I}_{1}}=\ln \dfrac{\sec x}{\sqrt{\dfrac{1}{{{e}^{2}}}+1}}\] .
From equation (4), we have \[I={{I}_{1}}+{{I}_{2}}\] .
Putting, \[{{I}_{1}}=\ln \dfrac{\sec x}{\sqrt{\dfrac{1}{{{e}^{2}}}+1}}\] and \[{{I}_{2}}=\ln (cosx)\] in \[I={{I}_{1}}+{{I}_{2}}\] , we get,
\[\begin{align}
  & I={{I}_{1}}+{{I}_{2}} \\
 & \Rightarrow I=\ln \dfrac{\sec x}{\sqrt{\dfrac{1}{{{e}^{2}}}+1}}+\ln \left( \dfrac{1}{\sec x} \right) \\
 & \Rightarrow I=\ln \left( \dfrac{\sec x}{\sqrt{\dfrac{1}{{{e}^{2}}}+1}}.\dfrac{1}{\sec x} \right) \\
 & \Rightarrow I=\ln \left( \dfrac{e}{\sqrt{1+{{e}^{2}}}} \right) \\
 & \Rightarrow I=\ln e-\ln (\sqrt{1+{{e}^{2}}}) \\
 & \Rightarrow I=1-1.06 \\
 & \Rightarrow I=-0.06 \\
\end{align}\]
Hence, there is no correct option.

Note: In this question, one may take the same upper and lower limits of definite integral even after assumption. Like for \[{{I}_{1}}=\int\limits_{\dfrac{1}{e}}^{\tan x}{\dfrac{tdt}{1+{{t}^{2}}}}\] , we have assumed \[P=1+{{t}^{2}}\] . After transforming one can write it as \[{{I}_{1}}=\dfrac{1}{2}\int\limits_{\dfrac{1}{e}}^{\tan x}{\dfrac{1}{P}}dP\] , which is wrong. Here, the upper and lower limits are the same as before. We need to change the upper and lower limits because we have transformed the expression and earlier limits were given for t. We have to find the new upper and lower limits of P. Therefore, our transformed expression should look like, \[{{I}_{1}}=\dfrac{1}{2}\int\limits_{1+\dfrac{1}{{{e}^{2}}}}^{{{\sec }^{2}}x}{\dfrac{1}{P}}dP\] .