Integration of root tanx

Integration of root tan x is little complex. This can be solve using integration by substitution method . The formula is given by

$\int \sqrt {tan x} \; dx = \frac {1}{\sqrt 2} \tan^{-1} (\frac {tan x -1}{\sqrt {2tan x}}) + \frac {1}{2\sqrt 2} \ln \frac {(tan x + 1 -\sqrt {2tan x})}{(tan x + 1 +\sqrt {2tan x})} + C$

Proof of integration of root tanx

Let $t^2 = \tan x$
$2tdt =\sec^2 x dx$
Now
$\sec^2 x = 1 + \tan^2 x$
Therefore
$2t dt = (1+ t^4) dx $
$dx = \frac {2t dt}{1+ t^4}$

So,

$= \int \frac {2t^2}{1+ t^4} \; dt $

$= \int \frac {t^2 +1 + t^2 -1 }{1+ t^4} \; dt $

$\int [\frac {t^2 + 1}{1+ t^4} + \frac {t^2 – 1}{1+ t^4}] \; dt $

$= \int \frac {t^2 + 1}{1+ t^4} \; dt + \int \frac {t^2 – 1}{1+ t^4} \; dt$

$= \int \frac {1 + 1/t^2}{t^2+ 1/t^2} \; dt + \int \frac {1 – 1/t^2}{t^2+ 1/t^2} \; dt$

$= \int \frac {1 + 1/t^2}{(t – 1/t)^2 +2} \; dt + \int \frac {1 – 1/t^2}{(t +1/t)^2 -2} \; dt$

Let $u =t – 1/t$ then $du = (1+ 1/t^2) dt$
Let $v =t + 1/t$ then $du = (1- 1/t^2) dt$

Therefore

$= \int \frac {1}{u^2 +2} \; du + \int \frac {1 }{v^2 -2} \; dv$

Now

Now
$\int \frac {1}{x^2 + a^2} dx = \frac {1}{a} \tan ^{-1} (\frac {x}{a}) + C$
Proof
Put $x =a tan \theta$ then $dx= a sec^2 \theta d\theta$
Therefore
$\int \frac {1}{x^2 + a^2} dx$
$=\int \frac {asec^2 \theta}{a^2 tan^2 \theta + a^2} d\theta$
$=\frac {1}{a} \int d\theta= \frac {1}{a} \theta + C = \frac {1}{a} \tan ^{-1} (\frac {x}{a}) + C$

$\int \frac {1}{x^2 – a^2} dx = \frac {1}{2a} ln |\frac {x-a}{x+a}| + C$
Proof
$\frac {1}{x^2 – a^2} =\frac {1}{2a}[ \frac {1}{x-a} – \frac {1}{x+a}]$
So
$\int \frac {1}{x^2 – a^2} dx $
$=\frac {1}{2a}[ \int \frac {1}{x-a} dx – \int \frac {1}{x+a}]$
$= \frac {1}{2a}[ln |x-a| – ln |x+a| + C$
$=\frac {1}{2a} ln |\frac {x-a}{x+a}| + C$

Therefore the above integral becomes

\[
=\frac {1}{\sqrt 2} tan^{-1} \frac {u}{\sqrt 2} + \frac {1}{2\sqrt 2} ln |\frac {v-\sqrt 2}{v+ \sqrt }| + C
\]

Now Substituting back the u and v

\[
=\frac {1}{\sqrt 2} tan^{-1} \frac {t^2 -1}{\sqrt 2 t} + \frac {1}{2\sqrt 2} ln |\frac {t^2 + 1-\sqrt 2 t}{t^2+ 1 + \sqrt t }| + C
\]

Now Substituting back the t

$\int \sqrt {tan x} \; dx = \frac {1}{\sqrt 2} \tan^{-1} (\frac {tan x -1}{\sqrt {2tan x}}) + \frac {1}{2\sqrt 2} \ln \frac {(tan x + 1 -\sqrt {2tan x})}{(tan x + 1 +\sqrt {2tan x})} + C$

Definite Integration of root tanx

Generally we calculate the definite integral by finding the indefinite integral and then finding the difference, but we can use definite integral rules to simplify that. Let see how

Example 1

$I = \int_{0}^{\pi/2} \sqrt {tan x} \; dx$ -(1)

Now we know that

$\int_{0}^{a} f(x) \; dx =\int_{0}^{a} f(a-x) \; dx $

Therefore

$I=\int_{0}^{\pi/2} \sqrt {tan (\pi/2 – x} \; dx= \int_{0}^{\pi/2} \sqrt {cot x} \; dx$ -(2)

Adding (1) and (2)

$2I= \int_{0}^{\pi/2} (\sqrt {tan x} + \sqrt {cot x}) \; dx$

$2I = \int_{0}^{\pi/2} \frac {sin x + cos x}{\sqrt {sinx cos x}} \; dx$

$2I = \sqrt 2 \int_{0}^{\pi/2} \frac {sin x + cos x}{\sqrt {2sinx cos x}} \; dx$

$2I = \sqrt 2 \int_{0}^{\pi/2} \frac {sin x + cos x}{\sqrt {1 -(sinx – cosx)^2}} \; dx$

Taking $u=sin x – cos x$
$du =( cos x + sin x) dx$

Therefore,

$2I = \sqrt 2 \int_{-1}^{1} \frac {1}{\sqrt {1-u^2}} \;du $
or

$2I = 2 \sqrt 2 \int_{0}^{1} \frac {1}{\sqrt {1-u^2}} \;du $
$I = \sqrt 2 \int_{0}^{1} \frac {1}{\sqrt {1-u^2}} \;du $

Now

$\int \frac {1}{\sqrt {a^2 – x^2}} dx =  \sin ^{-1} (\frac {x}{a}) + C$

Therefore

$I = \sqrt 2 [ \pi/2 – 0]= \frac {\pi}{\sqrt 2} $