The integration of cot^2x is equal to -(cot x +x)+C. In this post, we will learn how to find the integral of cot^2x (cot square x). We will use the following two formulas:

- ∫cosec
^{2}x dx = -cot x+C - ∫dx = x+C.

## Find Integration of cot^{2}x

Question: Find the integration of cot^{2} x?That is, find ∫cot ^{2} x dx. |

*Answer:* The integration of cot^{2}x is given by the formula ∫cot^{2}x dx = -(cotx +x)+C.

**Explanation:**

Recall the following trigonometric formula:

cosec^{2} x =1+cot^{2} x

∴ cot^{2} x = cosec^{2} x -1

Therefore, integrating both sides we obtain that

∫cot^{2}x dx = ∫(cosec^{2} x-1) dx +C

⇒ ∫cot^{2}x dx = ∫cosec^{2} x dx – ∫dx +C

⇒ ∫cot^{2}x dx = -cotx – x+C by the above two formulas.

So the integration of cot^2x (cot square x) is equal to -(cotx +x)+C where C denotes an arbitrary integral constant.

**ALSO READ:**

Integration of tan^{2}x | The integration of tan^{2}x is ∫tan^{2}x dx = tanx -x+C. |

Integration of sin^{2}x | The integration of sin^{2}x is ∫sin^{2}x dx = x/2-(sin2x)/4+C |

Integration of cos^{2}x | The integration of cos^{2}x is ∫cos^{2}x dx = x/2+(sin2x)/4+C |

## Definite Integral of cot^{2}x

**Question:** Find the definite integral of cot^{2}x from π/4 to π/2, that is,

find $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^2 x dx$

**Solution: **

From the above, we have ∫cot^{2}x dx = -cot x – x +C. Therefore, we get that

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^2 x dx$

= [-cot x -x+C]$_{\frac{\pi}{4}}^{\frac{\pi}{2}}$

= [-cot(π/2) – π/2 +C] – [-cot(π/4) – π/4 +C]

= -0 – π/2 +C +1 +π/4 -C as the value of cot(π/2) is 0 and the value of cot(π/4) is 1.

= 1- π/4

= (4-π)/4.

Thus, the definite integral of cot^{2}x from π/4 to π/2 is equal to (4-π)/4.

Have You Read These Integrals?

## FAQs

**Q1: What is the integration of cot ^{2}x?**

**Answer:** The integration cot square x is as follows: ∫cot^{2}x dx = -cotx – x +C, where C is an integral constant.