question-proposed-MN-july-1970-0-pi-4-tanx-ln-1-tan-2-x-dx-my-solution-followed-

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 \dots \dots .

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 \dots