What is the integral from 0 to pi/9 of xtan^2(3x) ?

The integral from 0 to pi/9 of xtan^2(3x) is (-sqrt(3)pi^2+18pi-18sqrt(3)ln(2))/(162sqrt(3))

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integral from 0 to pi/9 of xtan^2(3x)

\int_{0}^{\frac{π}{9}} x tan^{2} (3 x) d x

\frac{- 3 π ^{2} + 18 π - 18 3 ln ( 2 )}{162 3}

Decimal Notation

0.06359 \dots

Solution steps

\int_{0}^{\frac{π}{9}} x tan^{2} (3 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{9}} x (- 1 + sec^{2} (3 x)) d x

Apply u-substitution

: \int_{0}^{\frac{π}{3}} \frac{u tan ^{2} ( u )}{9} d u

= \int_{0}^{\frac{π}{3}} \frac{u tan ^{2} ( u )}{9} d u

Take the constant out:

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

= \frac{1}{9} \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}{9} [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}{9} [u (- u + tan (u)) - (- \frac{u ^{2}}{2} - ln ∣ cos (u) ∣)]_{0}^{\frac{π}{3}}

Simplify

\frac{1}{9} [u (- u + tan (u)) - (- \frac{u ^{2}}{2} - ln ∣ cos (u) ∣)]_{0}^{\frac{π}{3}} : \frac{1}{9} [\frac{- u ^{2} + 2 u tan ( u ) + 2 ln ∣ cos ( u ) ∣}{2}]_{0}^{\frac{π}{3}}

= \frac{1}{9} [\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}{9} \cdot \frac{- 3 π ^{2} + 18 π - 18 3 ln ( 2 )}{18 3}

Simplify

= \frac{- 3 π ^{2} + 18 π - 18 3 ln ( 2 )}{162 3}

\frac{d}{d x} (\frac{ln ( sin ( x ^{2} ) + 1 )}{f ( x )})

y = x^{2} - 2, (5, 23)

x \to 1 + lim (x^{\frac{1}{1 - x}})

\int \frac{2 x}{1 - x ^{2}} d x

\int (x^{2} + 1) e^{x} d x

What is the integral from 0 to pi/9 of xtan^2(3x) ?
The integral from 0 to pi/9 of xtan^2(3x) is (-sqrt(3)pi^2+18pi-18sqrt(3)ln(2))/(162sqrt(3))