What is the integral of 1/(2sqrt(tan(x))) ?

The integral of 1/(2sqrt(tan(x))) is

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integral of 1/(2sqrt(tan(x)))

\int \frac{1}{2 tan ( x )} d x

- 2 (- \frac{1}{8} (ln 2 cot (x) + 22 cot (x) + 2 - 2 arctan (2 cot (x) + 1)) + \frac{1}{8} (ln 2 cot (x) - 22 cot (x) + 2 + 2 arctan (2 cot (x) - 1))) + C

Solution steps

\int \frac{1}{2 tan ( x )} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{tan ( x )} d x

Rewrite using trig identities

= \frac{1}{2} \cdot \int cot (x) d x

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

= \frac{1}{2} (- \int \frac{2 v}{4 + v ^{2}} d v)

Take the constant out:

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

= \frac{1}{2} (- 2 \cdot \int \frac{v}{4 + v ^{2}} d v)

Apply u-substitution

: \int \frac{2 w ^{2}}{4 + w ^{4}} d w

= \frac{1}{2} (- 2 \cdot \int \frac{2 w ^{2}}{4 + w ^{4}} d w)

Take the constant out:

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

= \frac{1}{2} (- 2 \cdot 2 \cdot \int \frac{w ^{2}}{4 + w ^{4}} d w)

Take the partial fraction of

\frac{w ^{2}}{4 + w ^{4}} : - \frac{w}{4 ( w ^{2} + 2 w + 2 )} + \frac{w}{4 ( w ^{2} - 2 w + 2 )}

= \frac{1}{2} (- 2 \cdot 2 \cdot \int - \frac{w}{4 ( w ^{2} + 2 w + 2 )} + \frac{w}{4 ( w ^{2} - 2 w + 2 )} d w)

Apply the Sum Rule:

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

= \frac{1}{2} (- 2 \cdot 2 (- \int \frac{w}{4 ( w ^{2} + 2 w + 2 )} d w + \int \frac{w}{4 ( w ^{2} - 2 w + 2 )} d w))

\int \frac{w}{4 ( w ^{2} + 2 w + 2 )} d w = \frac{1}{8} (ln w^{2} + 2 w + 2 - 2 arctan (w + 1))

\int \frac{w}{4 ( w ^{2} - 2 w + 2 )} d w = \frac{1}{8} (ln w^{2} - 2 w + 2 + 2 arctan (w - 1))

= \frac{1}{2} (- 2 \cdot 2 (- \frac{1}{8} (ln w^{2} + 2 w + 2 - 2 arctan (w + 1)) + \frac{1}{8} (ln w^{2} - 2 w + 2 + 2 arctan (w - 1))))

Substitute back

= \frac{1}{2} (- 2 \cdot 2 (- \frac{1}{8} (ln (2 cot (x))^{2} + 2 2 cot (x) + 2 - 2 arctan (2 cot (x) + 1)) + \frac{1}{8} (ln (2 cot (x))^{2} - 2 2 cot (x) + 2 + 2 arctan (2 cot (x) - 1))))

Simplify

\frac{1}{2} (- 2 \cdot 2 (- \frac{1}{8} (ln (2 cot (x))^{2} + 2 2 cot (x) + 2 - 2 arctan (2 cot (x) + 1)) + \frac{1}{8} (ln (2 cot (x))^{2} - 2 2 cot (x) + 2 + 2 arctan (2 cot (x) - 1)))) : - 2 (- \frac{1}{8} (ln 2 cot (x) + 22 cot (x) + 2 - 2 arctan (2 cot (x) + 1)) + \frac{1}{8} (ln 2 cot (x) - 22 cot (x) + 2 + 2 arctan (2 cot (x) - 1)))

= - 2 (- \frac{1}{8} (ln 2 cot (x) + 22 cot (x) + 2 - 2 arctan (2 cot (x) + 1)) + \frac{1}{8} (ln 2 cot (x) - 22 cot (x) + 2 + 2 arctan (2 cot (x) - 1)))

Add a constant to the solution

= - 2 (- \frac{1}{8} (ln 2 cot (x) + 22 cot (x) + 2 - 2 arctan (2 cot (x) + 1)) + \frac{1}{8} (ln 2 cot (x) - 22 cot (x) + 2 + 2 arctan (2 cot (x) - 1))) + C

\frac{s ^{2} + 3 s + 5}{s ^{3} ( s + 2 )}

\frac{d y}{d x} = x

n = 0 \sum \infty \frac{2}{n ^{2}}

\int_{- 4}^{2} (x + 4) d x

y^{2} = 4 x, y = x^{2}

What is the integral of 1/(2sqrt(tan(x))) ?
The integral of 1/(2sqrt(tan(x))) is