What is the integral from 0 to pi/8 of xtan^2(2x) ?

The integral from 0 to pi/8 of xtan^2(2x) is 1/4 (pi/4-(pi^2)/(32)-1/2 ln(2))

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integral from 0 to pi/8 of xtan^2(2x)

\int_{0}^{\frac{π}{8}} x tan^{2} (2 x) d x

\frac{1}{4} (\frac{π}{4} - \frac{π ^{2}}{32} - \frac{1}{2} ln (2))

Decimal Notation

0.03259 \dots

Solution steps

\int_{0}^{\frac{π}{8}} 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}{8}}

= [x (- x + \frac{1}{2} tan (2 x)) - \int \frac{1}{2} (- 2 x + tan (2 x)) d x]_{0}^{\frac{π}{8}}

\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}{8}}

Simplify

[x (- x + \frac{1}{2} tan (2 x)) - \frac{1}{2} (- x^{2} - \frac{1}{2} ln ∣ cos (2 x) ∣)]_{0}^{\frac{π}{8}} : [\frac{1}{4} (- 2 x^{2} + 2 x tan (2 x) + ln ∣ cos (2 x) ∣)]_{0}^{\frac{π}{8}}

= [\frac{1}{4} (- 2 x^{2} + 2 x tan (2 x) + ln ∣ cos (2 x) ∣)]_{0}^{\frac{π}{8}}

Compute the boundaries:

\frac{1}{4} (\frac{π}{4} - \frac{π ^{2}}{32} - \frac{1}{2} ln (2))

= \frac{1}{4} (\frac{π}{4} - \frac{π ^{2}}{32} - \frac{1}{2} ln (2))

\int \frac{1}{4 u ^{2} + 9 - 41 u} d u

- sin (2 t)

\int \frac{1}{5 + 4 cos ( x )} d x

n = 6 \sum \infty \frac{5 ^{n}}{1 4 ^{n}}

\frac{d}{d x} (\frac{x}{x - 10})

What is the integral from 0 to pi/8 of xtan^2(2x) ?
The integral from 0 to pi/8 of xtan^2(2x) is 1/4 (pi/4-(pi^2)/(32)-1/2 ln(2))