question proposed MN july 1970 ∫_0 ^(π/4) tanx(ln(1+tan^2 x))dx my solution followed


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Question Number 111157 by mathdave last updated on 02/Sep/20

$q u e s t i o n p r o p o s e d M N j u l y 1970$ $\int_{0}^{\frac{π}{4}} \tan x (\ln (1 + \tan^{2} x)) d x$ $m y s o l u t i o n f o l l o w e d$
	Answered by mathdave last updated on 02/Sep/20

	$s i n c e 1 + \tan^{2} x = \sec^{2} x$ $l e t I = \int_{0}^{\frac{π}{4}} \tan x \ln (\sec^{2} x) d x = 2 \int_{0}^{\frac{π}{2}} \tan x \ln (\sec x) d x$ $l e t u = \ln (\sec x), d u = \tan x d x a n d d x = \frac{1}{\tan x} d u$ $I = 2 \int_{0}^{\frac{π}{4}} \tan x \times u \times \frac{1}{\tan x} d u = 2 \int_{0}^{\frac{π}{4}} u d u$ $I = 2 {[\frac{u^{2}}{2}]}_{0}^{\frac{π}{4}} = 2 {[\frac{\ln^{2} (\cos x)}{2}]}_{0}^{\frac{π}{4}} = [0 - \ln^{2} (\frac{1}{\sqrt{2}})] = \frac{1}{4} \ln^{2} (2)$ $∵ \int_{0}^{\frac{π}{4}} \tan x (\ln (1 + \tan^{2} x)) d x = \frac{1}{4} \ln^{2} (2)$ $m a t h d a v e (02 / 09 / 2020)$
	Commented by mnjuly1970 last updated on 02/Sep/20

	$t h a n k y o u s o m u c h s i r . . . . . . .$
	Commented by mathdave last updated on 02/Sep/20

	$y o u a r e w e l c o m e$
	Answered by mathmax by abdo last updated on 02/Sep/20

	$I = \int_{0}^{\frac{π}{4}} tanx (\ln (1 + \tan^{2} x)) dx changement tanx = t give$ $I = \int_{0}^{1} tln (1 + t^{2}) \frac{dt}{1 + t^{2}} =_{by parts} {[\frac{1}{2} \ln (1 + t^{2}) \ln (1 + t^{2})]}_{0}^{1}$ $- \int_{0}^{1} \frac{1}{2} \ln (1 + t^{2}) \frac{2 t}{1 + t^{2}} dt = \frac{1}{2} \ln^{2} (2) - \int_{0}^{1} \frac{tln (1 + t^{2})}{1 + t^{2}} dt$ $= \frac{1}{2} \ln^{2} (2) - I \Rightarrow 2 I = \frac{1}{2} \ln^{2} (2) \Rightarrow ★ I = \frac{1}{4} \ln^{2} (2) ★$
	Commented by mnjuly1970 last updated on 03/Sep/20

	$v e r y n i c e t h a n k y o u m a s t e r . . .$
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