Z-0-pi-2-arctan-sin-x-dx-0-pi-4-arcsin-tan-x-dx-

Question Number 135174 by bemath last updated on 11/Mar/21

Z = \int_{0}^{π / 2} \arctan (\sin x) dx + \int_{0}^{π / 4} \arcsin (\tan x) dx

Answered by john_santu last updated on 11/Mar/21

l e t Z_{1} = \int_{0}^{π / 2} \arctan (\sin x) d x

s e t t n g \sin x = q \Rightarrow Z_{1} = \int_{0}^{1} \frac{\arctan (q)}{\sqrt{1 - q^{2}}} d q

Z_{1} = {(\arctan (q) . \arcsin (q)]}_{0}^{1} - \int_{0}^{1} \frac{\arcsin q}{1 + q^{2}} d q

Z_{1} = \frac{π^{2}}{8} - \int_{0}^{1} \frac{\arcsin q}{1 + q^{2}} d q

l e t Z_{2} = \int_{0}^{π / 4} \arcsin (\tan x) d x

s e t t i n g \tan x = q

Z_{2} = \int_{0}^{1} \frac{\arcsin (q)}{1 + q^{2}} d q

N o w w e g e t Z = Z_{1} + Z_{2}

Z = \frac{π^{2}}{8} - \int_{0}^{1} \frac{\arcsin (q)}{1 + q^{2}} d q + \int_{0}^{1} \frac{\arcsin (q)}{1 + q^{2}} d q

Z = \frac{π^{2}}{8} ∙