I = ∫ _0 ^(π/2) ((1/(ln (tan r))) + (1/(1−tan r)) ) dr


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 120064 by bemath last updated on 29/Oct/20

$I = \int \underset{0}{\overset{\frac{π}{2}}{}} (\frac{1}{\ln (\tan r)} + \frac{1}{1 - \tan r}) d r$
	Answered by mnjuly1970 last updated on 29/Oct/20
	$I=^(∫_a ^( b) f(x)dx=∫_a ^( b) f(a+b−x)dx) ∫_0 ^( (π/2)) {(1/(−ln(tan(r))))+((tan(r))/(tan(r)−1))}dr =∫_0 ^( (π/2)) ((−1)/(ln(tan(r)))) +1−(1/(1−tan(r)))dr =−I+(π/2) 2I=(π/2) ⇒ I=(π/4) ✓ m.n.1970...$
	$I \overset{\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x}{=} \int_{0}^{\frac{π}{2}} {\frac{1}{- l n (t a n (r))} + \frac{t a n (r)}{t a n (r) - 1}} d r$ $= \int_{0}^{\frac{π}{2}} \frac{- 1}{l n (t a n (r))} + 1 - \frac{1}{1 - t a n (r)} d r$ $= - I + \frac{π}{2}$ $2 I = \frac{π}{2} \Rightarrow I = \frac{π}{4} ✓$ $m . n . 1970 . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum