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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2} tan^{2} (u) + tan (u) cot (u) + \frac{1}{2} cot^{2} (u) d u

= \int \frac{1}{2} tan^{2} (u) + tan (u) cot (u) + \frac{1}{2} cot^{2} (u) 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 \frac{1}{2} tan^{2} (u) d u + \int tan (u) cot (u) d u + \int \frac{1}{2} cot^{2} (u) d u

\int \frac{1}{2} tan^{2} (u) d u = \frac{1}{2} (- u + tan (u))

\int tan (u) cot (u) d u = u

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

= \frac{1}{2} (- u + tan (u)) + u + \frac{1}{2} (- u - cot (u))

Substitute back

u = 2 x

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

Simplify

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

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

Add a constant to the solution

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

\frac{6 x}{( 3 - 2 x ) ^{3}}

\frac{d}{d x} ((6 x + 5)^{2})

y^{^{''}} + 16 y = 24 sin (4 t)

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

f (x) = - 3 x

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