
Find the value of \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\left( {\sqrt {\cot x} } \right)}}{{\left[ {\sqrt {\cot x} + \sqrt {\tan x} } \right]}}} dx\]
A. \[1\]
B. \[ - 1\]
C. \[\dfrac{\pi }{2}\]
D. \[\dfrac{\pi }{4}\]
Answer
216.6k+ views
Hint: The question is given in \[\cot x\] which we will be changing into \[\tan x\] to get two equations which will help us to solve the question in the easiest way.
Formula Used: The formula we will be using is \[\left[ {\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {\left( {a - x} \right)dx} } \right]\]
Complete step by step answer
Let us assume the equation given in the question as (i) equation .
\[I = \] \[\] \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\left( {\sqrt {\cot x} } \right)}}{{\left[ {\sqrt {\cot x} + \sqrt {\tan x} } \right]}}} dx\] ----- (i)
Now we will be changing \[\cot x\] into \[\tan x\] by substituting \[x\] as \[\left( {\dfrac{\pi }{2} - x} \right)\] by applying the formula
\[I = \]\[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\cot \left( {\dfrac{\pi }{2} - x} \right)} }}{{\left[ {\sqrt {\cot \left( {\dfrac{\pi }{2} - x} \right)} + \sqrt {\tan \left( {\dfrac{\pi }{2} - x} \right)} } \right]}}} dx\]
\[\] \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\left( {\sqrt {\tan x} } \right)}}{{\left[ {\sqrt {\tan x} + \sqrt {\cot x} } \right]}}} dx\] ---- (ii)
Now , Adding equation (i) and (ii), we will get
\[I + I = 2I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\cot x} + \sqrt {\tan x} }}{{\left[ {\sqrt {\tan x} + \sqrt {\cot x} } \right]}}} dx\] \[\]
\[2I = \int\limits_0^{\dfrac{\pi }{2}} {1dx} \]
Integrating R.H.S. of the above equation and we get
\[2I = {\left[ x \right]^{\dfrac{\pi }{2}}}_0\]
Substitute the upper and lower value, we get
\[2I = \dfrac{\pi }{2} - 0\]
Dividing both sides of the above equation by 2, we get
\[I = \dfrac{\pi }{4}\]
Hence the answer is D which is \[\dfrac{\pi }{4}\].
Note: This is the easiest way of solving this question because here we have used the basic formula of integration where we have changed \[\cot x\] to \[\tan x\] and vice versa to get two equations which directly helped us to get the values of the integration.
Formula Used: The formula we will be using is \[\left[ {\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {\left( {a - x} \right)dx} } \right]\]
Complete step by step answer
Let us assume the equation given in the question as (i) equation .
\[I = \] \[\] \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\left( {\sqrt {\cot x} } \right)}}{{\left[ {\sqrt {\cot x} + \sqrt {\tan x} } \right]}}} dx\] ----- (i)
Now we will be changing \[\cot x\] into \[\tan x\] by substituting \[x\] as \[\left( {\dfrac{\pi }{2} - x} \right)\] by applying the formula
\[I = \]\[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\cot \left( {\dfrac{\pi }{2} - x} \right)} }}{{\left[ {\sqrt {\cot \left( {\dfrac{\pi }{2} - x} \right)} + \sqrt {\tan \left( {\dfrac{\pi }{2} - x} \right)} } \right]}}} dx\]
\[\] \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\left( {\sqrt {\tan x} } \right)}}{{\left[ {\sqrt {\tan x} + \sqrt {\cot x} } \right]}}} dx\] ---- (ii)
Now , Adding equation (i) and (ii), we will get
\[I + I = 2I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\cot x} + \sqrt {\tan x} }}{{\left[ {\sqrt {\tan x} + \sqrt {\cot x} } \right]}}} dx\] \[\]
\[2I = \int\limits_0^{\dfrac{\pi }{2}} {1dx} \]
Integrating R.H.S. of the above equation and we get
\[2I = {\left[ x \right]^{\dfrac{\pi }{2}}}_0\]
Substitute the upper and lower value, we get
\[2I = \dfrac{\pi }{2} - 0\]
Dividing both sides of the above equation by 2, we get
\[I = \dfrac{\pi }{4}\]
Hence the answer is D which is \[\dfrac{\pi }{4}\].
Note: This is the easiest way of solving this question because here we have used the basic formula of integration where we have changed \[\cot x\] to \[\tan x\] and vice versa to get two equations which directly helped us to get the values of the integration.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
Understanding Electromagnetic Waves and Their Importance

Geostationary and Geosynchronous Satellites Explained

Inertial and Non-Inertial Frame of Reference Explained

Clemmensen and Wolff Kishner Reductions Explained for JEE & NEET

JEE Main 2023 January 29th Shift 2 Physics Question Paper with Answer Keys and Solutions

Current Loop as a Magnetic Dipole: Concept, Derivation, and Examples

Other Pages
Amortization Calculator – Loan Schedule, EMI & Table

JEE Main Correction Window 2026 Session 1 Dates Announced - Edit Form Details, Dates and Link

Alpha, Beta, and Gamma Decay Explained for JEE & NEET

Devuthani Ekadashi 2025: Correct Date, Shubh Muhurat, Parana Time & Puja Vidhi

Quadratic Equation Questions with Solutions & PDF Practice Sets

Photosynthesis explained for students

