What is the integral from 0 to pi/6 of xtan^2(2x) ?

The integral from 0 to pi/6 of xtan^2(2x) is 1/4 (pi/(sqrt(3))-(pi^2)/(18)-ln(2))

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integral from 0 to pi/6 of xtan^2(2x)

\int_{0}^{\frac{π}{6}} x tan^{2} (2 x) d x

\frac{1}{4} (\frac{π}{3} - \frac{π ^{2}}{18} - ln (2))

Decimal Notation

0.14308 \dots

Solution steps

\int_{0}^{\frac{π}{6}} x tan^{2} (2 x) d x

Apply Integration By Parts

: [x (- x + \frac{1}{2} tan (2 x)) - \int \frac{1}{2} (- 2 x + tan (2 x)) d x]_{0}^{\frac{π}{6}}

= [x (- x + \frac{1}{2} tan (2 x)) - \int \frac{1}{2} (- 2 x + tan (2 x)) d x]_{0}^{\frac{π}{6}}

\int \frac{1}{2} (- 2 x + tan (2 x)) d x = \frac{1}{2} (- x^{2} - \frac{1}{2} ln ∣ cos (2 x) ∣)

= [x (- x + \frac{1}{2} tan (2 x)) - \frac{1}{2} (- x^{2} - \frac{1}{2} ln ∣ cos (2 x) ∣)]_{0}^{\frac{π}{6}}

Simplify

[x (- x + \frac{1}{2} tan (2 x)) - \frac{1}{2} (- x^{2} - \frac{1}{2} ln ∣ cos (2 x) ∣)]_{0}^{\frac{π}{6}} : [\frac{1}{4} (- 2 x^{2} + 2 x tan (2 x) + ln ∣ cos (2 x) ∣)]_{0}^{\frac{π}{6}}

= [\frac{1}{4} (- 2 x^{2} + 2 x tan (2 x) + ln ∣ cos (2 x) ∣)]_{0}^{\frac{π}{6}}

Compute the boundaries:

\frac{1}{4} (\frac{π}{3} - \frac{π ^{2}}{18} - ln (2))

= \frac{1}{4} (\frac{π}{3} - \frac{π ^{2}}{18} - ln (2))

x \to 0 lim (e^{3 x} - 1 - 3 x)

\frac{1}{4 x ^{4}}

\frac{d y}{d x} = 4 (x^{2} + 1), x (\frac{π}{4}) = 1

\int (8 x^{3} + \frac{1}{x ^{2}}) d x

y = 2^{- x}, (- 1, 2)

What is the integral from 0 to pi/6 of xtan^2(2x) ?
The integral from 0 to pi/6 of xtan^2(2x) is 1/4 (pi/(sqrt(3))-(pi^2)/(18)-ln(2))