What is the integral from 0 to 1 of 2tan(pi/4 y) ?

The integral from 0 to 1 of 2tan(pi/4 y) is (4ln(2))/pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of 2tan(pi/4 y)

\int_{0}^{1} 2 tan (\frac{π}{4} y) d y

\frac{4 ln ( 2 )}{π}

Decimal

0.88254 \dots

Solution steps

\int_{0}^{1} 2 tan (\frac{π}{4} y) d y

Take the constant out:

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

= 2 \cdot \int_{0}^{1} tan (\frac{π}{4} y) d y

Rewrite using trig identities

= 2 \cdot \int_{0}^{1} \frac{sin ( \frac{π y}{4} )}{cos ( \frac{π y}{4} )} d y

Apply u-substitution

: \int_{1}^{\frac{2}{2}} - \frac{4}{π u} d u

= 2 \cdot \int_{1}^{\frac{2}{2}} - \frac{4}{π u} d u

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

= 2 (- \int_{\frac{2}{2}}^{1} - \frac{4}{π u} d u)

Take the constant out:

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

= 2 (- (- \frac{4}{π} \cdot \int_{\frac{2}{2}}^{1} \frac{1}{u} d u))

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- (- \frac{4}{π} [ln ∣ u ∣]_{\frac{2}{2}}^{1}))

Simplify

2 (- (- \frac{4}{π} [ln ∣ u ∣]_{\frac{2}{2}}^{1})) : \frac{8}{π} [ln ∣ u ∣]_{\frac{1}{2}}^{1}

= \frac{8}{π} [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

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

= \frac{8}{π} \cdot \frac{1}{2} ln (2)

Simplify

= \frac{4 ln ( 2 )}{π}

\int 16 y d y

\int 2 x (x^{2} + 4)^{6} d x

\int e^{- s t} t e^{4 t} d t

y^{^{'}} = \frac{1}{t y}

g (x) = \frac{1}{5 x ^{2} + 6 x}

What is the integral from 0 to 1 of 2tan(pi/4 y) ?
The integral from 0 to 1 of 2tan(pi/4 y) is (4ln(2))/pi