What is the integral of (tan(x)+1/(tan(x)))^2 ?

The integral of (tan(x)+1/(tan(x)))^2 is -2cot(2x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (tan(x)+1/(tan(x)))^2

\int (tan (x) + \frac{1}{tan ( x )})^{2} d x

- 2 cot (2 x) + C

Solution steps

\int (tan (x) + \frac{1}{tan ( x )})^{2} d x

Rewrite using trig identities

= \int \frac{( sin ^{2} ( x ) + cos ^{2} ( x ) ) ^{2}}{cos ^{2} ( x ) sin ^{2} ( x )} d x

Expand

\frac{( sin ^{2} ( x ) + cos ^{2} ( x ) ) ^{2}}{cos ^{2} ( x ) sin ^{2} ( x )} : \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 2 + \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

= \int \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 2 + \frac{cos ^{2} ( x )}{sin ^{2} ( x )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{sin ^{2} ( x )}{cos ^{2} ( x )} d x + \int 2 d x + \int \frac{cos ^{2} ( x )}{sin ^{2} ( x )} d x

\int \frac{sin ^{2} ( x )}{cos ^{2} ( x )} d x = tan (x) - x

\int 2 d x = 2 x

\int \frac{cos ^{2} ( x )}{sin ^{2} ( x )} d x = - cot (x) - x

= tan (x) - x + 2 x - cot (x) - x

Simplify

tan (x) - x + 2 x - cot (x) - x : - 2 cot (2 x)

= - 2 cot (2 x)

Add a constant to the solution

= - 2 cot (2 x) + C

What is the integral of (tan(x)+1/(tan(x)))^2 ?
The integral of (tan(x)+1/(tan(x)))^2 is -2cot(2x)+C