Question

Statement – I: The value of the integral $\int_{\pi /6}^{\pi /3} {\dfrac{{{\text{dx}}}}{{1 + \sqrt {{\text{tan x}}} }}} $is equal to$\dfrac{\pi }{6}$.
Statement – II: $\int_{\text{a}}^{\text{b}} {{\text{f}}\left( {\text{x}} \right){\text{dx}}} = \int_{\text{a}}^{\text{b}} {{\text{f}}\left( {{\text{a + b - x}}} \right){\text{dx}}} $
$
  {\text{A}}{\text{. Statement - I is true, Statement - II is true; Statement - II is not a correct explanation for Statement - I}} \\
  {\text{B}}{\text{. Statement - I is true, Statement - II is false}} \\
  {\text{C}}{\text{. Statement - I is false, Statement - II is true}} \\
  {\text{D}}{\text{. Statement - I is true, Statement - II is true; Statement - II is a correct explanation for Statement - I}} \\
$

Answer 1

Hint: In order to determine the correct answer we compute statements I and II first and then verify which of the options holds true. We solve the given statements using the formula of property of definite integrals, which says$\int_{\text{a}}^{\text{b}} {{\text{f}}\left( {\text{x}} \right){\text{dx}}} = \int_{\text{a}}^{\text{b}} {{\text{f}}\left( {{\text{a + b - x}}} \right){\text{dx}}} $. We refer to the concept of definite integrals to solve and verify the both statements.

Complete step-by-step solution:
Given data,
 $\int_{\pi /6}^{\pi /3} {\dfrac{{{\text{dx}}}}{{1 + \sqrt {{\text{tan x}}} }}} $
$\int_{\text{a}}^{\text{b}} {{\text{f}}\left( {\text{x}} \right){\text{dx}}} = \int_{\text{a}}^{\text{b}} {{\text{f}}\left( {{\text{a + b - x}}} \right){\text{dx}}} $
Let us consider the integral function given in statement I to be ‘X’
$ \Rightarrow {\text{X = }}\int_{\pi /6}^{\pi /3} {\dfrac{{{\text{dx}}}}{{1 + \sqrt {{\text{tan x}}} }}} {\text{ - - - - }}\left( 1 \right)$
Now according to the property of definite integrals we know that an integral function of the form $\int_{\text{a}}^{\text{b}} {{\text{f}}\left( {\text{x}} \right){\text{dx}}} $can be expressed as$\int_{\text{a}}^{\text{b}} {{\text{f}}\left( {{\text{a + b - x}}} \right){\text{dx}}} $.
I.e., $\int_{\text{a}}^{\text{b}} {{\text{f}}\left( {\text{x}} \right){\text{dx}}} = \int_{\text{a}}^{\text{b}} {{\text{f}}\left( {{\text{a + b - x}}} \right){\text{dx}}} $
This is Statement II itself, hence Statement – II is true.
Let us apply this property on equation – (1), we get
$\Rightarrow {\text{X = }}\int_{\pi /6}^{\pi /3} {\dfrac{{{\text{dx}}}}{{1 + \sqrt {{\text{tan }}\left( {\dfrac{\pi }{2} - {\text{x}}} \right)} }}} \\
   \Rightarrow {\text{X = }}\int_{\pi /6}^{\pi /3} {\dfrac{{{\text{dx}}}}{{1 + \sqrt {{\text{cot x}}} }}} \\
   \Rightarrow {\text{X = }}\int_{\pi /6}^{\pi /3} {\dfrac{{\sqrt {{\text{tan x}}} }}{{1 + \sqrt {{\text{tan x}}} }}{\text{dx}}} {\text{ - - - }}\left( 2 \right) \\ $
Adding equations (1) and (2) we get
$\Rightarrow 2{\text{X = }}\int_{\pi /6}^{\pi /3} {\left[ {\dfrac{{\sqrt {{\text{tan x}}} }}{{1 + \sqrt {{\text{tan x}}} }} + \dfrac{1}{{1 + \sqrt {{\text{tan x}}} }}} \right]{\text{dx}}} \\
   \Rightarrow 2{\text{X = }}\int_{\pi /6}^{\pi /3} {{\text{dx}}} \\
   \Rightarrow 2{\text{X = }}\left[ {\text{x}} \right]_{\pi /6}^{\pi /3} \\
   \Rightarrow 2{\text{X = }}\dfrac{\pi }{6} \\
   \Rightarrow {\text{X = }}\dfrac{\pi }{{12}} \\ $
Hence we get $\int_{\pi /6}^{\pi /3} {\dfrac{{{\text{dx}}}}{{1 + \sqrt {{\text{tan x}}} }}} = \dfrac{\pi }{{12}}$
Therefore the statement is false.
Hence we get Statement – I is false, Statement – II is true.
Option C is the correct answer.

Note: In order to solve this type of question the key is to know the concepts of integration and definite integrals. Once we identify the statement – II is an identity, it is true. Then we use the same to compute the answer of integral in statement I without even simplifying the integral. General knowledge of how to perform integration and substitute limits later is required to perform integration problems.