# Integral of square root tanx | Find ∫√tanx dx

The integral of root tanx is equal to

∫√tanx dx = $\dfrac{1}{\sqrt{2}}$ tan-1 ($\dfrac{\tan x-1}{\sqrt{2\tan x}}$) + $\dfrac{1}{2\sqrt{2}}$ log $|\dfrac{\tan x+1-\sqrt{2\tan x}}{\tan x+1+\sqrt{2\tan x}}|$ +C

where C denotes an integral constant. In this post, we will learn how to find the integral of square root of tanx.

## Integration of √tanx

Question: Find ∫√tanx dx.

Solution:

Let tanx=t2.

Differentiating w.r.t x, sec2x dx = 2tdt

∴ dx = $\dfrac{2t dt}{\sec^2x}$ = $\dfrac{2tdt}{1+\tan^2x}$ = $\dfrac{2tdt}{1+t^4}$.

So the given integral of square root of tanx can be written as

∫√tanx dx

= ∫$\dfrac{2t^2 dt}{1+t^4}$

= ∫$\dfrac{(t^2+1)+(t^2-1)}{1+t^4}dt$

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

Divide both the numerator and the denominator by t2.

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

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

= I1 + I2 (say).

That is, ∫√tanx dx = I1 + I2 …(∗)

Therefore, from (∗) we get that

∫√tanx dx = I1 + I2

= $\int \dfrac{du}{u^2+2} dt$ + $\int \dfrac{dv}{v^2-2} dt$

= $\dfrac{1}{\sqrt{2}}$ tan-1 ($\dfrac{u}{\sqrt{2}}$) + $\dfrac{1}{2\sqrt{2}}$ log $|\dfrac{v-\sqrt{2}}{v+\sqrt{2}}|$ +C

= $\dfrac{1}{\sqrt{2}}$ tan-1 ($\dfrac{t^2-1}{\sqrt{2}t}$) + $\dfrac{1}{2\sqrt{2}}$ log $|\dfrac{t^2+1-\sqrt{2}t}{t^2+1+\sqrt{2}t}|$ +C (putting the above values of u and v)

= $\dfrac{1}{\sqrt{2}}$ tan-1 ($\dfrac{\tan x-1}{\sqrt{2\tan x}}$) + $\dfrac{1}{2\sqrt{2}}$ log $|\dfrac{\tan x+1-\sqrt{2\tan x}}{\tan x+1+\sqrt{2\tan x}}|$ +C.

So the integral of square root of tanx is equal to $\dfrac{1}{\sqrt{2}}$ tan-1 ($\dfrac{\tan x-1}{\sqrt{2\tan x}}$) + $\dfrac{1}{2\sqrt{2}}$ log $|\dfrac{\tan x+1-\sqrt{2\tan x}}{\tan x+1+\sqrt{2\tan x}}|$ +C, where C is an integration constant.

Related Integrals:

Find ∫tanx dx

Find ∫tan2x dx

Evaluate ∫log(sinx) dx

Find ∫cotx dx

Find ∫sin3x cos2x dx

## FAQs

Q1: What is the integration of root tan x?

Answer: The integration of tan x is equal to ∫√tanx dx = $\dfrac{1}{\sqrt{2}}$ tan-1 ($\dfrac{\tan x-1}{\sqrt{2\tan x}}$) + $\dfrac{1}{2\sqrt{2}}$ log $|\dfrac{\tan x+1-\sqrt{2\tan x}}{\tan x+1+\sqrt{2\tan x}}|$ +C.