calculate-0-pi-4-sinx-ln-cosx-dx-

Question Number 35060 by math khazana by abdo last updated on 14/May/18

c a l c u l a t e \int_{0}^{\frac{π}{4}} s i n x l n (c o s x) d x

Commented by math khazana by abdo last updated on 15/May/18

l e t p u t I = \int_{0}^{\frac{π}{4}} s i n x l n (c o s x) d x c h a n g e m e n t

c o s x = t g i v e

I = \int_{1}^{\frac{\sqrt{2}}{2}} \sqrt{1 - t^{2}} l n (t) - \frac{d t}{\sqrt{1 - t^{2}}}

= \int_{\frac{\sqrt{2}}{2}}^{1} l n (t) d t = {[t l n (t) - t_{}]}_{\frac{\sqrt{2}}{2}}^{1}

= - 1 - \frac{\sqrt{2}}{2} l n (\frac{\sqrt{2}}{2}) + \frac{\sqrt{2}}{2}

= \frac{\sqrt{2}}{2} - 1 - \frac{\sqrt{2}}{2} {\frac{1}{2} l n (2) - l n (2)}

= \frac{\sqrt{2}}{2} - 1 + \frac{1}{4} \sqrt{2} l n (2) .

Answered by tanmay.chaudhury50@gmail.com last updated on 15/May/18

t = c o s x d t = - s i n x d x

\int_{1}^{\frac{1}{\sqrt{2}}} - l n t d t

I^{'} = \int l n t d t

= l n t \times t - t

s o I = - ∣ t l n t - t ∣_{1}^{\frac{1}{\sqrt{2}}}

= - [{\frac{1}{\sqrt{2}} l n (\frac{1}{\sqrt{2}}) - \frac{1}{\sqrt{2}}} - {1 . l n 1 - 1}]

= - [\frac{1}{\sqrt{2}} (- \frac{1}{2} l n 2) - \frac{1}{\sqrt{2}}} - {- 1}]

= - [(- \frac{1}{2 \sqrt{2}} l n 2 - \frac{1}{\sqrt{2}}} + 1]

= \frac{1}{2 \sqrt{2}} l n 2 + \frac{1}{\sqrt{2}} - 1

p l s c h e c k

Commented by math khazana by abdo last updated on 15/May/18

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