What is the integral from 0 to pi/(12) of xtan^2(4x) ?

The integral from 0 to pi/(12) of xtan^2(4x) is (-sqrt(3)pi^2+18pi-18sqrt(3)ln(2))/(288sqrt(3))

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/(12) of xtan^2(4x)

\int_{0}^{\frac{π}{12}} x tan^{2} (4 x) d x

\frac{- 3 π ^{2} + 18 π - 18 3 ln ( 2 )}{288 3}

Decimal Notation

0.03577 \dots

Solution steps

\int_{0}^{\frac{π}{12}} x tan^{2} (4 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{12}} x (- 1 + sec^{2} (4 x)) d x

Apply u-substitution

: \int_{0}^{\frac{π}{3}} \frac{u tan ^{2} ( u )}{16} d u

= \int_{0}^{\frac{π}{3}} \frac{u tan ^{2} ( u )}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int_{0}^{\frac{π}{3}} u tan^{2} (u) d u

Apply Integration By Parts

: [u (- u + tan (u)) - \int - u + tan (u) d u]_{0}^{\frac{π}{3}}

= \frac{1}{16} [u (- u + tan (u)) - \int - u + tan (u) d u]_{0}^{\frac{π}{3}}

\int - u + tan (u) d u = - \frac{u ^{2}}{2} - ln ∣ cos (u) ∣

= \frac{1}{16} [u (- u + tan (u)) - (- \frac{u ^{2}}{2} - ln ∣ cos (u) ∣)]_{0}^{\frac{π}{3}}

Simplify

\frac{1}{16} [u (- u + tan (u)) - (- \frac{u ^{2}}{2} - ln ∣ cos (u) ∣)]_{0}^{\frac{π}{3}} : \frac{1}{16} [\frac{- u ^{2} + 2 u tan ( u ) + 2 ln ∣ cos ( u ) ∣}{2}]_{0}^{\frac{π}{3}}

= \frac{1}{16} [\frac{- u ^{2} + 2 u tan ( u ) + 2 ln ∣ cos ( u ) ∣}{2}]_{0}^{\frac{π}{3}}

Compute the boundaries:

\frac{- 3 π ^{2} + 18 π - 18 3 ln ( 2 )}{18 3}

= \frac{1}{16} \cdot \frac{- 3 π ^{2} + 18 π - 18 3 ln ( 2 )}{18 3}

Simplify

= \frac{- 3 π ^{2} + 18 π - 18 3 ln ( 2 )}{288 3}

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

\int_{1}^{3} \frac{1}{x ^{2} + 81} d x

\int_{0}^{1} (4 - 3 x^{2}) d x

\int_{0}^{y} ln (y) d y

What is the integral from 0 to pi/(12) of xtan^2(4x) ?
The integral from 0 to pi/(12) of xtan^2(4x) is (-sqrt(3)pi^2+18pi-18sqrt(3)ln(2))/(288sqrt(3))