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

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

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

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

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

Decimal

0.57234 \dots

Solution steps

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

Apply Integration By Parts

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

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

\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}{3}}

Simplify

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

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

Compute the boundaries:

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

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

\frac{d}{d x} (ln (\frac{1 + x}{1 - x}))

5 x + 4 cos (x)

y = 5 e^{4 x} cos (5 x)

\frac{s}{( s ^{2} + 4 ) ^{2}}

f (x) = 6 sin (x) + 4 cos (x), at x = 0

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