Question

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 1

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.