What is the integral of sqrt(1+cot^2(x)) ?

The integral of sqrt(1+cot^2(x)) is ln|tan(x/2)|+C

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integral of sqrt(1+cot^2(x))

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

ln tan (\frac{x}{2}) + C

Solution steps

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

Rewrite using trig identities

= \int csc^{2} (x) d x

csc^{2} (x) = (csc (x)),

assuming

csc (x) \geq 0

= \int csc (x) d x

Apply u-substitution

: \int \frac{1}{u} d u

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = tan (\frac{x}{2})

= ln tan (\frac{x}{2})

Add a constant to the solution

= ln tan (\frac{x}{2}) + C

\int_{0}^{5} ∣ 15 - 5 t ∣ d t

\frac{d}{d x} (f (x) x^{^{'}} (x))

\frac{\partial}{\partial x} (ln (8 x^{2} + 3 y^{2} + 3))

\int \frac{10 x ^{4} - 9 x ^{2}}{2 x ^{5} - 3 x ^{3} + 14} d x

y^{^{'''}} + 3 y^{^{''}} + 3 y^{^{'}} + y = 0

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