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

The integral of tan(x/2) is -2ln|cos(x/2)|+C

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integral of tan(x/2)

\int tan (\frac{x}{2}) d x

- 2 ln cos (\frac{x}{2}) + C

Solution steps

\int tan (\frac{x}{2}) d x

Rewrite using trig identities

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

Apply u-substitution

: \int - \frac{2}{u} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 2 ln ∣ u ∣

Substitute back

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

= - 2 ln cos (\frac{x}{2})

Add a constant to the solution

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

\frac{\partial}{\partial x} (x^{2} + y cos (x y))

\int 4 sec^{4} (x) tan (x) d x

\frac{d}{d x} (\frac{x ^{5}}{5} + \frac{1}{12 x ^{3}})

x d y - (x^{2} - y^{2} + y) \cdot d x = 0

y = \frac{x + 2}{x - 2}, (3, 5)