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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate :$\int \sqrt \tan x \;dx\;$

(Hint:Put$\;tan x=t^2$)

$\begin{array}{1 1} (A) \large\frac{1}{\sqrt 2}\normalsize \tan^{-1}\bigg(\large\frac{\tan x+\cot x}{\sqrt 2}\bigg)+\frac{1}{2\sqrt 2}\normalsize\log\begin{vmatrix} \large\frac{\tan x+\cot x-\sqrt 2}{\tan x+\cot x+\sqrt 2}\end{vmatrix}+c \\ (B) \large\frac{1}{\sqrt 2}\normalsize \tan^{-1}\bigg(\large\frac{\tan x-\cot x}{\sqrt 2}\bigg)+\frac{1}{2\sqrt 2}\normalsize\log\begin{vmatrix} \large\frac{\tan x+\cot x-\sqrt 2}{\tan x+\cot x-\sqrt 2}\end{vmatrix}+c \\ (C) \large\frac{1}{\sqrt 2}\normalsize \tan^{-1}\bigg(\large\frac{\tan x-\cot x}{\sqrt 2}\bigg)+\frac{1}{2\sqrt 2}\normalsize\log\begin{vmatrix} \large\frac{\tan x-\cot x-\sqrt 2}{\tan x-\cot x+\sqrt 2}\end{vmatrix}+c \\ (D) \large\frac{1}{\sqrt 2}\normalsize \tan^{-1}\bigg(\large\frac{\tan x-\cot x}{\sqrt 2}\bigg)+\frac{1}{2\sqrt 2}\normalsize\log\begin{vmatrix} \large\frac{\tan x+\cot x-\sqrt 2}{\tan x+\cot x+\sqrt 2}\end{vmatrix}+c \end{array} $

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1 Answer

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Toolbox:
  • If $f(x)$ is substituted by $f(t)$,then $f'(x)dx=f'(t) dt.$
  • Hence $f(x) dx=f(t) dt$
  • $\int\large\frac{dx}{x^2+a^2}=\frac{1}{a}\normalsize \tan^{-1}\big(\frac{x}{a}\big)+c$
Step 1:
Let $I=\int \sqrt{\tan x}dx.$
Put $\tan x=t^2.$.Hence on differentiating with respect to x we get,
$\sec^2x dx=2t dt\Rightarrow dx=\large\frac{2t dt}{\sec^2x}$
But $\sec^2x=1+\tan ^2x=1+t^2$
$dx=\large\frac{2t dt}{1+t^4}$
$I=\int t.\large\frac{2t}{1+t^4}\normalsize dt.$
$\;\;=2\int \large\frac{t^2}{1+t^4}\normalsize dt.$
Add and subtract I to the numerator
$I=\int \large\frac{t^2+1+t^2-1}{1+t^4}\normalsize dt.$
$I=I_1+I_2$
Step 2:
Consider $I_1=\int\large\frac{t^2+1}{t^4+1}\normalsize dt$.This can be written as
$\int\large\frac{t^2(1+\large\frac{1}{t^2})}{t^2(t^2+\frac{1}{t^2})}dt=\int\large\frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}\normalsize dt.$
But $t^2+\large\frac{1}{t^2}=(t-\frac{1}{t})^2\normalsize +2.$
Let $(t-\frac{1}{t})=u.$
$(1+\frac{1}{t^2})dt=du.$
$I_1=\int\large\frac{1+\frac{1}{t^2}}{(t-\frac{1}{t})^2+2}=\int\frac{du}{u^2+(\sqrt 2)^2}$
This is of the form $\int\large\frac{dx}{1+x^2}=\frac{1}{a}\tan^{-1}\big(\frac{x}{a}\big)+c.$
Hence $x=u$ and $a=\sqrt 2$
On integrating we get,
$\large\frac{1}{\sqrt 2}\normalsize \tan^{-1}\big(\frac{u}{\sqrt 2}\big)+c.$
Now substituting back for u we get,
$\large\frac{1}{\sqrt 2}\normalsize \tan^{-1}\large\frac{(t-\frac{1}{t})}{\sqrt 2}+c.$
But $t=\tan x$
$I_1=\large\frac{1}{\sqrt 2}\tan^{-1}\bigg(\frac{\tan x-\cot x}{\sqrt 2}\bigg)+c$
Step 3:
Next consider $I_2=\int \large\frac{t^2-1}{1+t^4}\normalsize dt.$
Divide the numerator and denominator by $t^2$
$I_2=\int \large\frac{ \frac{t^2}{t^2}-\frac{1}{t^2}}{\frac{1}{t^2}+\frac{t^4}{t^2}}\normalsize dt.$
$\quad=\int\large \frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt.$
$\quad=\int \large\frac{1-\frac{1}{t^2}}{(t+\frac{1}{t})^2-(\sqrt 2)^2}\normalsize dt.$
Now substituting this we get,
$\int \large\frac{du}{u^2-(\sqrt 2)^2}\normalsize du.$
This is of the form $\int \large\frac{dx}{x^2-a^2}$
$\;\;=\frac{1}{2a}\log \mid\large\frac{x-a}{x+a}\mid\normalsize+c$
Step 4:
Here $x=u$ and $a=\sqrt 2$
Hence on integrating we get,
$I_2=\large\frac{1}{2\sqrt 2}\log\mid\frac{u-\sqrt 2}{u+\sqrt 2}\mid+c$
Substituting for u we get,
$I_2=\large\frac{1}{2\sqrt 2}\log\mid\frac{t+\frac{1}{t}-\sqrt 2}{t+\frac{1}{t}+\sqrt 2}\mid+c$
But $t=\tan x$
$I_2=\large\frac{1}{2\sqrt 2}\log\mid\frac{\tan x+cot x-\sqrt 2}{\tan x+\cot x+\sqrt 2}\mid+c$
Now $I=I_1+I_2$
On combining $I_1$ and $I_2$ we get,
$I=\large\frac{1}{\sqrt 2}\normalsize \tan^{-1}\bigg(\large\frac{\tan x-\cot x}{\sqrt 2}\bigg)+\frac{1}{2\sqrt 2}\normalsize\log\begin{vmatrix} \large\frac{\tan x+\cot x-\sqrt 2}{\tan x+\cot x+\sqrt 2}\end{vmatrix}+c$
answered Apr 23, 2013 by sreemathi.v
edited Feb 6, 2014 by rvidyagovindarajan_1
 
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