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

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

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integral of (tan(x)+cot(x))^2

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

- 2 cot (2 x) + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2} + 1}{u ^{2}} d u

= \int \frac{u ^{2} + 1}{u ^{2}} d u

\frac{u ^{2} + 1}{u ^{2}} = \frac{1}{u ^{2}} + 1

= \int \frac{1}{u ^{2}} + 1 d u

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{u ^{2}} = u^{- 2}

= \int u^{- 2} + 1 d u

Apply the Sum Rule:

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

= \int u^{- 2} d u + \int 1 d u

\int u^{- 2} d u = - \frac{1}{u}

\int 1 d u = u

= - \frac{1}{u} + u

Substitute back

u = tan (x)

= - \frac{1}{tan ( x )} + tan (x)

Simplify

- \frac{1}{tan ( x )} + tan (x) : - 2 cot (2 x)

= - 2 cot (2 x)

Add a constant to the solution

= - 2 cot (2 x) + C

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

\frac{\partial}{\partial x} (e^{4 x yz - 9})

\frac{2 x + 6 x ^{2}}{( 1 + 6 x ) ^{2}}

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