# How to Integrate Square Root cotx | Find ∫√cotx dx

The integral of root cotx is equal to

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

where C is an arbitrary integral constant. Here we will learn how to integrate square root of cotx.

## Integral of √cotx

Question: Find ∫√cotx dx.

Solution:

Let cotx=t2.

Differentiating w.r.t x, -cosec2x dx = 2tdt

∴ dx = $\dfrac{-2t dt}{\text{cosec}^2x}$ = $\dfrac{-2tdt}{1+\cot^2x}$ = $\dfrac{-2tdt}{1+t^4}$.

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

∫√cotx dx

= ∫$\dfrac{-2t^2 dt}{1+t^4}$

= – ∫$\dfrac{(t^2+1)+(t^2-1)}{1+t^4}dt$

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

Dividing both the numerator and the denominator by t2, the above integral

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

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

= I1 + I2 (say).

So we get ∫√cotx dx = I1 + I2 …(∗)

Now, from (∗),

∫√cotx 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 ($\dfrac{\cot x-1}{\sqrt{2\cot x}}$) – $\dfrac{1}{2\sqrt{2}}$ log $|\dfrac{\cot x+1-\sqrt{2\cot x}}{\cot x+1+\sqrt{2\cot x}}|$ +C.

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

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