Integrate Square Root of tan x

Integration is a vital topic in calculus and plays an important role in solving real-life mathematical problems. Today, we’ll tackle a slightly tricky one, i.e. to integrate square root of tan x.

Breaking Down the Integral

$\int \sqrt{\tan x}\,dx$

The given integral, may seem complex at first, but with the right substitution, it becomes manageable.

Let’s start with substitution method to make it simple.

Simplifying the Integral: $\int \sqrt{\tan x}\,dx$

Solution

Let

$I=\int \sqrt{\tan x}\,dx$

Put

$\tan x=t^2$

Then

$\sec^2x\,dx=2t\,dt$

$dx=\frac{2t\,dt}{\sec^2x}$

Using $\sec^2x=1+\tan^2x$ , we get

$dx=\frac{2t\,dt}{1+t^4}$

Now, we need to do an important step here to solve the problem, i.e. breaking down the integral into two integrals, as I have done in the image below:

Substituting in the integral:

$\begin{aligned} I &=\int \sqrt{t^2}\cdot \frac{2t}{1+t^4}\,dt \\ &=\int \frac{2t^2}{1+t^4}\,dt \end{aligned}$

Now write

$2t^2=(t^2+1)+(t^2-1)$

Therefore,

$\begin{aligned} I &=\int \frac{t^2+1}{1+t^4}\,dt+\int \frac{t^2-1}{1+t^4}\,dt \\ &=I_1+I_2 \end{aligned}$ …(1)

Evaluate integeral of √tan x — Watch it on YouTube

Key Insight to Integrate Square Root of tan x: Special Integrals

The integral can be simplified further using special integrals and substitution method of integration again. After splitting the given integral into two integrals, we need to solve them one-by-one by the method of completing the perfect square in the denominator.

Evaluating $I_1$

$I_1=\int \frac{t^2+1}{1+t^4}\,dt$

Divide numerator and denominator by $t^2$ :

$\begin{aligned} I_1 &=\int \frac{1+\frac1{t^2}}{t^2+\frac1{t^2}}\,dt \\ &=\int \frac{1+\frac1{t^2}} {t^2+\frac1{t^2}-2+2}\,dt \end{aligned}$

Using

$t^2+\frac1{t^2}-2=\left(t-\frac1t\right)^2$

we get

$I_1= \int \frac{1+\frac1{t^2}} {\left(t-\frac1t\right)^2+2}\,dt$

Put

$u=t-\frac1t$

Then

$du=\left(1+\frac1{t^2}\right)dt$

Hence,

$\begin{aligned} I_1 &=\int \frac{du}{u^2+2} \\ &=\int \frac{du}{u^2+(\sqrt2)^2} \\ &=\frac1{\sqrt2}\tan^{-1}\left(\frac{u}{\sqrt2}\right) \end{aligned}$

Substituting back:

$I_1= \frac1{\sqrt2} \tan^{-1} \left( \frac{t-\frac1t}{\sqrt2} \right) + C_1$ …(2)

Moreover, we need to be careful before making the substitution again in both the integrals separately, as you can see in the next image below. (I suggest you to refer to the full PDF (Free Download button at the end) for step-by-step hand-written solution or the YouTube video here)

Integration by substitution — Watch it on YouTube

Evaluating $I_2$

$I_2=\int \frac{t^2-1}{1+t^4}\,dt$

Divide numerator and denominator by $t^2$ :

$\begin{aligned} I_2 &=\int \frac{1-\frac1{t^2}}{t^2+\frac1{t^2}}\,dt \\ &=\int \frac{1-\frac1{t^2}} {t^2+\frac1{t^2}+2-2}\,dt \end{aligned}$

Using

$t^2+\frac1{t^2}+2=\left(t+\frac1t\right)^2$

we get

$I_2= \int \frac{1-\frac1{t^2}} {\left(t+\frac1t\right)^2-2}\,dt$

Put

$v=t+\frac1t$

Then

$dv=\left(1-\frac1{t^2}\right)dt$

Hence,

$\begin{aligned} I_2 &=\int \frac{dv}{v^2-2} \\ &=\int \frac{dv}{v^2-(\sqrt2)^2} \\ &=\frac1{2\sqrt2} \log\left| \frac{v-\sqrt2}{v+\sqrt2} \right| \end{aligned}$

Substituting back:

$I_2= \frac1{2\sqrt2} \log\left| \frac{ t+\frac1t-\sqrt2 }{ t+\frac1t+\sqrt2 } \right| + C_2$ …(3)

Final Answer: Integration of root tan x

Since $I=I_1+I_2$ , therefore from (1), (2) and (3), we get

$\boxed{ \begin{aligned} \int \sqrt{\tan x}\,dx &= \frac1{\sqrt2} \tan^{-1} \left( \frac{ t-\frac1t }{\sqrt2} \right) \\ &\quad+ \frac1{2\sqrt2} \log\left| \frac{ t+\frac1t-\sqrt2 }{ t+\frac1t+\sqrt2 } \right| +C \end{aligned} }$

where

$t=\sqrt{\tan x}$

Don’t forget to put back the values of u, v & t (or any other letters/symbols you may have used for the substitution!) everywhere in both the integrals. And the following image has the final answer:

Integrate square root of tan x — Watch it on YouTube

Practice Question

1. Don’t miss to integrate $\sqrt{\tan x} + \sqrt{\cot x}$ also. Want to check your answer, watch a full video here?

Conclusion

Integrating square root of tan x is challenging but becomes straightforward with the right substitution and methodical steps. For a complete explanation, including intermediate steps and tricks, download the full PDF solution below. (You can also watch a how-to-solve video on YouTube!)

INTEGRATION OF SQUARE ROOT OF TAN X (pdf)Download

Integrate Square Root of tan x

Breaking Down the Integral

Simplifying the Integral: $\int \sqrt{\tan x}\,dx$