Integrate $\int {\sqrt {\tan x} dx} $ with respect to $x$.
Answer
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Hint: In this question, we have been asked to integrate $\int {\sqrt {\tan x} dx} $. At first, assume $\tan x = {t^2}$. Then, differentiate the equation and find the value of dx. Put them in the given equation. Then, divide the question into parts by adding and subtracting 1 in the numerator. Both the parts have the same steps. Solve each part individually. First, divide the numerator and the denominator with ${t^2}$. Then, manipulate the denominator in such a way that the denominator has a value which can be substituted for a letter, whose differentiated value is in the numerator only. After you have changed the denominator, put it equal to a letter and differentiate it. You will find its value in the numerator. Substitute it and apply the relevant formula. Repeat the same process with the other part. Add both the parts together. You will have your answer.
Complete step-by-step solution:
We have been asked to integrate $\int {\sqrt {\tan x} dx} $ with respect to $x$. The solution to this question is a little lengthy and you should be aware about certain trigonometric identities to find the answer. Let us see how it is done.
Firstly, we will try to remove the square root. For this purpose, we will assume a certain value of $\tan x$.
Let $\tan x = {t^2}$. (We will assume it is equal to the square so when it comes under root, it will be simplified.)
$ \Rightarrow \tan x = {t^2}$
Differentiating both the sides with respect to $x$
$ \Rightarrow {\sec ^2}xdx = 2tdt$
Now, we have to find the value of dx in terms of t. So, we will put ${\sec ^2}x = 1 + {\tan ^2}x$. Now, we know that $\tan x = {t^2}$
Putting this in ${\sec ^2}x = 1 + {\tan ^2}x$, we get
$ \Rightarrow {\sec ^2}x = 1 + {\operatorname{t} ^4}$
Now we have the following value of dx –
$ \Rightarrow dx = \dfrac{{2tdt}}{{1 + {t^4}}}$
Putting this in the given equation,
$ \Rightarrow \int {\dfrac{{2{t^2}dt}}{{1 + {t^4}}}} $
Now, we will separate the numerator and add and subtract 1 in the numerator.
$ \Rightarrow \int {\dfrac{{{t^2} + 1 + {t^2} - 1dt}}{{1 + {t^4}}}} $
Now, we will separate the entire question into 2 parts.
$ \Rightarrow \int {\dfrac{{1 + {t^2}}}{{1 + {t^4}}}dt + \int {\dfrac{{{t^2} - 1}}{{1 + {t^4}}}dt} } $ …. (1)
Let $\int {\dfrac{{1 + {t^2}}}{{1 + {t^4}}}dt = {I_1}} $ and $\int {\dfrac{{{t^2} - 1}}{{1 + {t^4}}}dt = {I_2}} $.
Now, we will solve both the parts individually. Starting with part 1,
$ \Rightarrow {I_1} = \int {\dfrac{{1 + {t^2}}}{{1 + {t^4}}}dt} $
Dividing numerator and denominator by ${t^2}$,
$ \Rightarrow {I_1} = \int {\dfrac{{\dfrac{1}{{{t^2}}} + 1}}{{\dfrac{1}{{{t^2}}} + {t^2}}}dt} $
Now, we will manipulate the denominator in such a way that we get something which can be substituted for a letter and we get their differentiation in numerator.
We know that ${\left( {t - \dfrac{1}{t}} \right)^2} = {t^2} + \dfrac{1}{{{t^2}}} - 2$. We can write it as - ${\left( {t - \dfrac{1}{t}} \right)^2} + 2 = {t^2} + \dfrac{1}{{{t^2}}}$. Substituting this in the equation,
$ \Rightarrow {I_1} = \int {\dfrac{{\dfrac{1}{{{t^2}}} + 1}}{{{{\left( {t - \dfrac{1}{t}} \right)}^2} + 2}}dt} $ …. (2)
Now, we will assume $t - \dfrac{1}{t} = u$ and we will differentiate both the sides with respect to t.
$ \Rightarrow \left( {1 + \dfrac{1}{{{t^2}}}} \right)dt = du$
Putting this in equation (2),
$ \Rightarrow {I_1} = \int {\dfrac{{du}}{{{{\left( u \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}}}} $
Using the formula - $\int {} \dfrac{{dx}}{{{a^2} + {x^2}}} = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + C$,
$ \Rightarrow {I_1} = \dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\dfrac{u}{{\sqrt 2 }} + {C_1}$
Substituting the value of $u$
$ \Rightarrow {I_1} = \dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{{{t^2} - 1}}{{\sqrt 2 t}}} \right) + {C_1}$
Substituting $\tan x = {t^2}$
$ \Rightarrow {I_1} = \dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{{\tan x - 1}}{{\sqrt {2\tan x} }}} \right) + {C_1}$
Now, we will move towards second part.
$ \Rightarrow {I_2} = \int {\dfrac{{{t^2} - 1}}{{1 + {t^4}}}dt} $
In this part also, we will follow the same steps as followed above.
Dividing numerator and denominator by ${t^2}$
$ \Rightarrow {I_2} = \int {\dfrac{{1 - \dfrac{1}{{{t^2}}}}}{{\dfrac{1}{{{t^2}}} + {t^2}}}dt} $
We know that ${\left( {t + \dfrac{1}{t}} \right)^2} = {t^2} + \dfrac{1}{{{t^2}}} + 2$
We can write it as - ${\left( {t + \dfrac{1}{t}} \right)^2} - 2 = {t^2} + \dfrac{1}{{{t^2}}}$
Substituting this in the equation,
$ \Rightarrow {I_2} = \int {\dfrac{{1 - \dfrac{1}{{{t^2}}}}}{{{{\left( {t + \dfrac{1}{t}} \right)}^2} - 2}}dt} $ …. (3)
Now, we will assume $t + \dfrac{1}{t} = v$ and we will differentiate both the sides with respect to t.
$ \Rightarrow \left( {1 - \dfrac{1}{{{t^2}}}} \right)dt = dv$
Putting this in equation (3),
$ \Rightarrow {I_2} = \int {\dfrac{{dv}}{{{{\left( v \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}}} $
Using the formula - $\int {\dfrac{{dx}}{{{a^2} - {x^2}}} = \dfrac{1}{{2a}}\log \left| {\dfrac{{x - a}}{{x + a}}} \right|} $,
$ \Rightarrow {I_2} = \dfrac{1}{{2\sqrt 2 }}\log \left| {\dfrac{{v - \sqrt 2 }}{{v + \sqrt 2 }}} \right| + {C_2}$
Now, put the value of $v$
$ \Rightarrow {I_2} = \dfrac{1}{{2\sqrt 2 }}\log \left| {\dfrac{{\dfrac{{{t^2} + 1}}{t} - \sqrt 2 }}{{\dfrac{{{t^2} + 1}}{t} + \sqrt 2 }}} \right| + {C_2}$
Simplifying,
$ \Rightarrow {I_2} = \dfrac{1}{{2\sqrt 2 }}\log \left| {\dfrac{{{t^2} + 1 - \sqrt 2 t}}{{{t^2} + 1 + \sqrt 2 t}}} \right| + {C_2}$
Substituting $\tan x = {t^2}$
$ \Rightarrow {I_2} = \dfrac{1}{{2\sqrt 2 }}\log \left| {\dfrac{{\tan x + 1 - \sqrt {2\tan x} }}{{\tan x + 1 + \sqrt {2\tan x} }}} \right| + {C_2}$
Now, for the final answer, we will add ${I_1}$ and ${I_2}$
$ \Rightarrow I = {I_1} + {I_2}$
$ \Rightarrow I = \dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{{\tan x - 1}}{{\sqrt {2\tan x} }}} \right) + \dfrac{1}{{2\sqrt 2 }}\log \left| {\dfrac{{\tan x + 1 - \sqrt {2\tan x} }}{{\tan x + 1 + \sqrt {2\tan x} }}} \right| + C$
Hence, this is our final answer.
Note: We have to remember that, in mathematics, an integral assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data. Integration is one of the two main operations of calculus. Its inverse operation of differentiation, is the other. The operation of integration, up to an additive constant, is the inverse of the operation of differentiation.
Complete step-by-step solution:
We have been asked to integrate $\int {\sqrt {\tan x} dx} $ with respect to $x$. The solution to this question is a little lengthy and you should be aware about certain trigonometric identities to find the answer. Let us see how it is done.
Firstly, we will try to remove the square root. For this purpose, we will assume a certain value of $\tan x$.
Let $\tan x = {t^2}$. (We will assume it is equal to the square so when it comes under root, it will be simplified.)
$ \Rightarrow \tan x = {t^2}$
Differentiating both the sides with respect to $x$
$ \Rightarrow {\sec ^2}xdx = 2tdt$
Now, we have to find the value of dx in terms of t. So, we will put ${\sec ^2}x = 1 + {\tan ^2}x$. Now, we know that $\tan x = {t^2}$
Putting this in ${\sec ^2}x = 1 + {\tan ^2}x$, we get
$ \Rightarrow {\sec ^2}x = 1 + {\operatorname{t} ^4}$
Now we have the following value of dx –
$ \Rightarrow dx = \dfrac{{2tdt}}{{1 + {t^4}}}$
Putting this in the given equation,
$ \Rightarrow \int {\dfrac{{2{t^2}dt}}{{1 + {t^4}}}} $
Now, we will separate the numerator and add and subtract 1 in the numerator.
$ \Rightarrow \int {\dfrac{{{t^2} + 1 + {t^2} - 1dt}}{{1 + {t^4}}}} $
Now, we will separate the entire question into 2 parts.
$ \Rightarrow \int {\dfrac{{1 + {t^2}}}{{1 + {t^4}}}dt + \int {\dfrac{{{t^2} - 1}}{{1 + {t^4}}}dt} } $ …. (1)
Let $\int {\dfrac{{1 + {t^2}}}{{1 + {t^4}}}dt = {I_1}} $ and $\int {\dfrac{{{t^2} - 1}}{{1 + {t^4}}}dt = {I_2}} $.
Now, we will solve both the parts individually. Starting with part 1,
$ \Rightarrow {I_1} = \int {\dfrac{{1 + {t^2}}}{{1 + {t^4}}}dt} $
Dividing numerator and denominator by ${t^2}$,
$ \Rightarrow {I_1} = \int {\dfrac{{\dfrac{1}{{{t^2}}} + 1}}{{\dfrac{1}{{{t^2}}} + {t^2}}}dt} $
Now, we will manipulate the denominator in such a way that we get something which can be substituted for a letter and we get their differentiation in numerator.
We know that ${\left( {t - \dfrac{1}{t}} \right)^2} = {t^2} + \dfrac{1}{{{t^2}}} - 2$. We can write it as - ${\left( {t - \dfrac{1}{t}} \right)^2} + 2 = {t^2} + \dfrac{1}{{{t^2}}}$. Substituting this in the equation,
$ \Rightarrow {I_1} = \int {\dfrac{{\dfrac{1}{{{t^2}}} + 1}}{{{{\left( {t - \dfrac{1}{t}} \right)}^2} + 2}}dt} $ …. (2)
Now, we will assume $t - \dfrac{1}{t} = u$ and we will differentiate both the sides with respect to t.
$ \Rightarrow \left( {1 + \dfrac{1}{{{t^2}}}} \right)dt = du$
Putting this in equation (2),
$ \Rightarrow {I_1} = \int {\dfrac{{du}}{{{{\left( u \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}}}} $
Using the formula - $\int {} \dfrac{{dx}}{{{a^2} + {x^2}}} = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + C$,
$ \Rightarrow {I_1} = \dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\dfrac{u}{{\sqrt 2 }} + {C_1}$
Substituting the value of $u$
$ \Rightarrow {I_1} = \dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{{{t^2} - 1}}{{\sqrt 2 t}}} \right) + {C_1}$
Substituting $\tan x = {t^2}$
$ \Rightarrow {I_1} = \dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{{\tan x - 1}}{{\sqrt {2\tan x} }}} \right) + {C_1}$
Now, we will move towards second part.
$ \Rightarrow {I_2} = \int {\dfrac{{{t^2} - 1}}{{1 + {t^4}}}dt} $
In this part also, we will follow the same steps as followed above.
Dividing numerator and denominator by ${t^2}$
$ \Rightarrow {I_2} = \int {\dfrac{{1 - \dfrac{1}{{{t^2}}}}}{{\dfrac{1}{{{t^2}}} + {t^2}}}dt} $
We know that ${\left( {t + \dfrac{1}{t}} \right)^2} = {t^2} + \dfrac{1}{{{t^2}}} + 2$
We can write it as - ${\left( {t + \dfrac{1}{t}} \right)^2} - 2 = {t^2} + \dfrac{1}{{{t^2}}}$
Substituting this in the equation,
$ \Rightarrow {I_2} = \int {\dfrac{{1 - \dfrac{1}{{{t^2}}}}}{{{{\left( {t + \dfrac{1}{t}} \right)}^2} - 2}}dt} $ …. (3)
Now, we will assume $t + \dfrac{1}{t} = v$ and we will differentiate both the sides with respect to t.
$ \Rightarrow \left( {1 - \dfrac{1}{{{t^2}}}} \right)dt = dv$
Putting this in equation (3),
$ \Rightarrow {I_2} = \int {\dfrac{{dv}}{{{{\left( v \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}}} $
Using the formula - $\int {\dfrac{{dx}}{{{a^2} - {x^2}}} = \dfrac{1}{{2a}}\log \left| {\dfrac{{x - a}}{{x + a}}} \right|} $,
$ \Rightarrow {I_2} = \dfrac{1}{{2\sqrt 2 }}\log \left| {\dfrac{{v - \sqrt 2 }}{{v + \sqrt 2 }}} \right| + {C_2}$
Now, put the value of $v$
$ \Rightarrow {I_2} = \dfrac{1}{{2\sqrt 2 }}\log \left| {\dfrac{{\dfrac{{{t^2} + 1}}{t} - \sqrt 2 }}{{\dfrac{{{t^2} + 1}}{t} + \sqrt 2 }}} \right| + {C_2}$
Simplifying,
$ \Rightarrow {I_2} = \dfrac{1}{{2\sqrt 2 }}\log \left| {\dfrac{{{t^2} + 1 - \sqrt 2 t}}{{{t^2} + 1 + \sqrt 2 t}}} \right| + {C_2}$
Substituting $\tan x = {t^2}$
$ \Rightarrow {I_2} = \dfrac{1}{{2\sqrt 2 }}\log \left| {\dfrac{{\tan x + 1 - \sqrt {2\tan x} }}{{\tan x + 1 + \sqrt {2\tan x} }}} \right| + {C_2}$
Now, for the final answer, we will add ${I_1}$ and ${I_2}$
$ \Rightarrow I = {I_1} + {I_2}$
$ \Rightarrow I = \dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{{\tan x - 1}}{{\sqrt {2\tan x} }}} \right) + \dfrac{1}{{2\sqrt 2 }}\log \left| {\dfrac{{\tan x + 1 - \sqrt {2\tan x} }}{{\tan x + 1 + \sqrt {2\tan x} }}} \right| + C$
Hence, this is our final answer.
Note: We have to remember that, in mathematics, an integral assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data. Integration is one of the two main operations of calculus. Its inverse operation of differentiation, is the other. The operation of integration, up to an additive constant, is the inverse of the operation of differentiation.
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