The integration of tan^2x is equal to ∫tan^{2}x dx = tan (x)-x. In this post, we will learn how to find the integral of tan square x. We will use the following two formulas:

- ∫sec
^{2}x dx =tan x+C - ∫dx = x+C

## Integration of tan^{2}x

**Question:** What is the integration of tan^{2} x? That is, find ∫tan^{2} x dx.

*Answer:* ∫tan^{2}x dx = tan(x) -x +C.

**Explanation:**

To find the integration of tan^{2}x, we will proceed as follows. At first, we will use the formula given below.

sec^{2} x =1+tan^{2} x

⇒ tan^{2} x = sec^{2} x -1

Therefore, we have that ∫tan^{2}x dx = ∫(sec^{2} x-1) dx

= ∫sec^{2} x dx – ∫dx

= tan x – x +C by the above formulas.

So the integration of tan^{2} x is equal to tan(x)-x+C where C is an integration constant.

## Definite Integral of tan^{2}x

**Question:** Find the definite integral $\int_0^{\frac{\pi}{4}} \tan^2 x dx$

*Answer: *

From the above, we know that the integration of tan square x is ∫tan^{2}x dx = tan x – x +C. Therefore, the given definite integral will be equal to

$\int_0^{\frac{\pi}{4}} \tan^2 x dx$

= $[\tan x -x]_0^{\frac{\pi}{4}}$

= $(\tan \frac{\pi}{4}- \frac{\pi}{4})$ $-(\tan 0 -0)$

= $(1- \frac{\pi}{4})-(0 -0)$ as the value of $\tan \frac{\pi}{4}$ is $1$ and the value of tan0 is 0.

= $1- \frac{\pi}{4}$.

Thus, the integral of tan^{2}x from 0 to π/4 is equal to 1-π/4.

**Also Read:**

Integration of log(sinx) from 0 to pi/2

Derivative & Integration of 1/root(x)

## FAQs

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

**Answer:** The integration formula of tan square x is as follows: ∫tan^{2}x dx = tan(x) -x +C.