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

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

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integral of (tan(θ)+cot(θ))^2

\int (tan (θ) + cot (θ))^{2} d θ

- 2 cot (2 θ) + C

Solution steps

\int (tan (θ) + cot (θ))^{2} d θ

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 (θ)

= - \frac{1}{tan ( θ )} + tan (θ)

Simplify

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

= - 2 cot (2 θ)

Add a constant to the solution

= - 2 cot (2 θ) + C

\int \frac{6}{x} + 6 x d x

(- 27 sin (3 x) - 27 x cos (3 x))^{^{'}}

f (x) = \frac{2 x}{x ^{2} + 1}, at x = - 1, - 1

\frac{2 e ^{- s}}{( s + \frac{1}{5} )}

y^{^{''}} - 5 y^{^{'}} - 14 y = 0

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