What is the integral from 0 to pi/4 of cotan(x) ?

The integral from 0 to pi/4 of cotan(x) is (ln(2)co)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi/4 of cotan(x)

\int_{0}^{\frac{π}{4}} co tan (x) d x

\frac{ln ( 2 ) co}{2}

Solution steps

\int_{0}^{\frac{π}{4}} co tan (x) d x

Take the constant out:

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

= co \cdot \int_{0}^{\frac{π}{4}} tan (x) d x

Rewrite using trig identities

= co \cdot \int_{0}^{\frac{π}{4}} \frac{sin ( x )}{cos ( x )} d x

Apply u-substitution

: \int_{1}^{\frac{2}{2}} - \frac{1}{u} d u

= co \cdot \int_{1}^{\frac{2}{2}} - \frac{1}{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= co (- \int_{\frac{2}{2}}^{1} - \frac{1}{u} d u)

Take the constant out:

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

= co (- (- \int_{\frac{2}{2}}^{1} \frac{1}{u} d u))

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= co (- (- [ln ∣ u ∣]_{\frac{2}{2}}^{1}))

Simplify

co (- (- [ln ∣ u ∣]_{\frac{2}{2}}^{1})) : co [ln ∣ u ∣]_{\frac{1}{2}}^{1}

= co [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

\frac{1}{2} ln (2)

= co \frac{1}{2} ln (2)

Simplify

= \frac{ln ( 2 ) co}{2}

\int_{0}^{z} z^{2} d z

\int_{0}^{2} (3 x^{2} - 1) d x

\int_{0}^{1} (4 - 4 t^{3}) d t

\int_{0}^{4} x sin (\frac{nπ x}{4}) d x

\int_{0}^{x} \frac{1}{1 + x + x ^{2}} d x

What is the integral from 0 to pi/4 of cotan(x) ?
The integral from 0 to pi/4 of cotan(x) is (ln(2)co)/2