
Find the value of \[\int {\left[ {\dfrac{{\left( {\sqrt {tan{\rm{ }}x}} \right)}}{{\left( {sin{\rm{ }}x{\rm{ }}cos{\rm{ }}x} \right)}}{\rm{ }}} \right]} {\rm{ }}dx\]. When \[{\rm{ }}x{\rm{ }} \ne {\rm{ }}\dfrac{{k\pi }}{2}{\rm{ and }}tan{\rm{ }}x{\rm{ }} > {\rm{ }}0\].
A. \[\left( {\dfrac{1}{{2\sqrt {tan{\rm{ }}x}}}{\rm{ }}} \right)\]
B. \[\sqrt {\left( {2{\rm{ }}tan{\rm{ }}x} \right)}\]
C. \[2{\rm{ }}\sqrt {\left( {tan{\rm{ }}x} \right)}\]
D. \[\sqrt {tan{\rm{ }}x}\]
Answer
161.1k+ views
Hint: We will proceed this problem by writing the trigonometric ratio \[\tan x\] in \[\sin x\] and \[\cos x\] form. Then we will put the inverse ratio of \[\cos x\] in numerator and combine the terms in obtained form. Then after using the substitution method of integration to obtain the solution of given function.
Formula used:
The trigonometric ratio \[\tan x = \dfrac{{\sin x}}{{\cos x}}\] and the integration of any constant A is \[\int {Adx = } \;Ax + c\], Where A and C are the constants. Also, to simplify the problem in later part we have substituted the simple variable in place of trigonometric variable.
Complete step by step solution:
The trigonometric ratio \[\tan x\] is written in \[\sin x\] and \[\cos x\] form,
\[\int {\dfrac{{\sqrt {\tan x} }}{{\sin x\cos x}}} dx = \int {\dfrac{{\sqrt {\sin x} }}{{\sin x\cos x\sqrt {\cos x} }}} dx\]
The inverse ratio of \[\cos x\] that is \[\sec x\] write in numerator we get,
\[ \Rightarrow \int {\dfrac{{\sec x\;dx}}{{\sqrt {\sin x} (\sqrt {\cos x} )}}} \]
Multiply and divide by \[\sec x\] we get,
\[ \Rightarrow \int {\dfrac{{{{\sec }^2}x}}{{\sqrt {\sin x} \sqrt {\cos x} \sec x}}} dx\]
Convert the \[\sec x\] in denominator in \[\cos x\] and evaluate we get,
\[ \Rightarrow \int {\dfrac{{{{\sec }^2}x}}{{\sqrt {\tan x} }}} dx\] ………… (1)
Now Substitute the value while letting \[\tan x = {z^2}\] , Where\[z\] is a supposed variable.
Then we get,
\[{\sec ^2}xdx = 2zdz\] ……….. (2)
Substitute the equation (2) in equation (1) we get,
\[ \Rightarrow \int {\dfrac{{2zdz}}{z}} = 2z + c\]
Obtain the required result by putting the original value in place of supposed value.
If \[\tan x = {z^2}\], then \[z = \sqrt {\tan x} \].
Therefore, we get
\[ \Rightarrow 2\sqrt {\tan x} + c\]
Hence option C is correct.
Note: This problem can also be proceeded by multiplying the numerator and denominator with \[\sqrt {\tan x} \] therefore obtaining \[\int {\dfrac{{{{\sec }^2}x}}{{\sqrt {\sin x} \sqrt {\cos x} \sec x}}} dx\] directly. Students may find difficultly in figuring out the variable which should be substituted in place of \[\tan x\], but this issue can be resolved by through practicing of this method. Also, we should not forget to substitute the original value in place of supposed value in end as here in end we put \[z = \sqrt {\tan x} \] in our final solution.
Formula used:
The trigonometric ratio \[\tan x = \dfrac{{\sin x}}{{\cos x}}\] and the integration of any constant A is \[\int {Adx = } \;Ax + c\], Where A and C are the constants. Also, to simplify the problem in later part we have substituted the simple variable in place of trigonometric variable.
Complete step by step solution:
The trigonometric ratio \[\tan x\] is written in \[\sin x\] and \[\cos x\] form,
\[\int {\dfrac{{\sqrt {\tan x} }}{{\sin x\cos x}}} dx = \int {\dfrac{{\sqrt {\sin x} }}{{\sin x\cos x\sqrt {\cos x} }}} dx\]
The inverse ratio of \[\cos x\] that is \[\sec x\] write in numerator we get,
\[ \Rightarrow \int {\dfrac{{\sec x\;dx}}{{\sqrt {\sin x} (\sqrt {\cos x} )}}} \]
Multiply and divide by \[\sec x\] we get,
\[ \Rightarrow \int {\dfrac{{{{\sec }^2}x}}{{\sqrt {\sin x} \sqrt {\cos x} \sec x}}} dx\]
Convert the \[\sec x\] in denominator in \[\cos x\] and evaluate we get,
\[ \Rightarrow \int {\dfrac{{{{\sec }^2}x}}{{\sqrt {\tan x} }}} dx\] ………… (1)
Now Substitute the value while letting \[\tan x = {z^2}\] , Where\[z\] is a supposed variable.
Then we get,
\[{\sec ^2}xdx = 2zdz\] ……….. (2)
Substitute the equation (2) in equation (1) we get,
\[ \Rightarrow \int {\dfrac{{2zdz}}{z}} = 2z + c\]
Obtain the required result by putting the original value in place of supposed value.
If \[\tan x = {z^2}\], then \[z = \sqrt {\tan x} \].
Therefore, we get
\[ \Rightarrow 2\sqrt {\tan x} + c\]
Hence option C is correct.
Note: This problem can also be proceeded by multiplying the numerator and denominator with \[\sqrt {\tan x} \] therefore obtaining \[\int {\dfrac{{{{\sec }^2}x}}{{\sqrt {\sin x} \sqrt {\cos x} \sec x}}} dx\] directly. Students may find difficultly in figuring out the variable which should be substituted in place of \[\tan x\], but this issue can be resolved by through practicing of this method. Also, we should not forget to substitute the original value in place of supposed value in end as here in end we put \[z = \sqrt {\tan x} \] in our final solution.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

JEE Advanced 2025 Notes

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
