find-0-pi-2-x-2-tan-2-x-dx-

Question Number 207565 by mathzup last updated on 18/May/24

f i n d \int_{0}^{\frac{π}{2}} \frac{x^{2}}{{t a n}^{2} x} d x

Answered by sniper237 last updated on 19/May/24

\int_{0}^{π / 2} \frac{x^{2} (1 - {s i n}^{2} x)}{{s i n}^{2} x} d x = \int_{0}^{π / 2} \frac{x^{2}}{{s i n}^{2} x} d x - \frac{π^{3}}{24}

\overset{P I}{=} {[- \frac{x^{2}}{t a n x}]}_{\to 0} + 2 \int_{0}^{π / 2} \frac{x}{t a n x} d x - \frac{π^{3}}{24}

\overset{P I}{=} 2 {[x l n (s i n x)]}_{\to 0} - 2 \int_{0}^{π / 2} l n (s i n x) d x - \frac{π^{3}}{24}

\overset{t = s i n x}{=} - 2 \int_{0}^{1} \frac{l n t}{\sqrt{1 - t^{2}}} d t - \frac{π^{3}}{24}

= - 2 \partial_{a} {(\int_{0}^{1} \frac{t^{a}}{\sqrt{1 - t^{2}}} d t)}_{0} - \frac{π^{3}}{24}

\overset{u = t^{2}}{=} - \partial_{a} (\int_{0}^{1} u^{\frac{a - 1}{2}} {(1 - u)}^{\frac{1}{2} - 1} d u) - \frac{π^{3}}{24}

= - \partial_{a} {(\frac{Γ (\frac{a + 1}{2}) Γ (\frac{1}{2})}{Γ (\frac{a}{2} + 1)})}_{0} - \frac{π^{3}}{24}

= - \frac{1}{2} \frac{Γ^{'} (1 / 2) Γ (1 / 2) - Γ^{'} (1) Γ {(1 / 2)}^{2}}{1^{2}} - \frac{π^{3}}{24}

= \frac{3 π γ - 6 \sqrt{π} Γ^{'} (1 / 2) - π^{3}}{24}

Answered by Berbere last updated on 19/May/24

x \to \frac{π}{2} - x

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

- \int_{0}^{\frac{π}{2}} {(\frac{π}{2} - x)}^{2} = A - B

B = - \frac{1}{3} {[{(\frac{π}{2} - x)}^{3}]}_{0}^{\frac{π}{2}} = \frac{π^{3}}{24}

A = {[\tan (x) {(\frac{π}{2} - x)}^{2}]}_{0}^{\frac{π}{2}} - 2 \int_{0}^{\frac{π}{2}} \tan (x) (\frac{π}{2} - x)

= - 2 \int_{0}^{\frac{π}{2}} \tan (x) (\frac{π}{2} - x) d x = {[2 l n (c o s (x)) (\frac{π}{2} - x)]}_{0}^{\frac{π}{2}}

- 2 \int_{0}^{\frac{π}{2}} l n (c o s (x)) d x = - 2 . - \frac{π}{2} l n (2) = π l n (2)

π l n (2) - \frac{π^{3}}{24}

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