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

The integral from 0 to pi/4 of xtan^2(x) is 1/2 (pi/2-(pi^2)/(16)-ln(2))

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integral from 0 to pi/4 of xtan^2(x)

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

\frac{1}{2} (\frac{π}{2} - \frac{π ^{2}}{16} - ln (2))

Decimal Notation

0.13039 \dots

Solution steps

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

Apply Integration By Parts

: [x (- x + tan (x)) - \int - x + tan (x) d x]_{0}^{\frac{π}{4}}

= [x (- x + tan (x)) - \int - x + tan (x) d x]_{0}^{\frac{π}{4}}

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

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

Simplify

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

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

Compute the boundaries:

\frac{1}{2} (\frac{π}{2} - \frac{π ^{2}}{16} - ln (2))

= \frac{1}{2} (\frac{π}{2} - \frac{π ^{2}}{16} - ln (2))

\int_{1}^{e} (ln (x))^{3} d x

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\int_{4}^{6} x d x

\int_{- 1}^{5} 4 x (1 - x^{2}) d x

\int_{- 1}^{2} 2 x^{2} d x

What is the integral from 0 to pi/4 of xtan^2(x) ?
The integral from 0 to pi/4 of xtan^2(x) is 1/2 (pi/2-(pi^2)/(16)-ln(2))