calculate ∫_0 ^(π/4) sinx ln(cosx)dx


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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
$let put I = ∫_0 ^(π/4) sinx ln(cosx)dx changement cosx =t give I = ∫_1 ^((√2)/2) (√(1−t^2 )) ln(t) −(dt/(√(1−t^2 ))) = ∫_((√2)/2) ^1 ln(t)dt = [ t ln(t) −t_ ]_((√2)/2) ^1 = −1 −((√2)/2)ln(((√2)/2)) +((√2)/2) =((√2)/2) −1 −((√2)/2){ (1/2)ln(2) −ln(2)} =((√2)/2) −1 +(1/4)(√2) ln(2) .$
$l e t p u t I = \int_{0}^{\frac{π}{4}} s i n x l n (c o s x) d x c h a n g e m e n t$ $c o s x = t g i v e$ $I = \int_{1}^{\frac{\sqrt{2}}{2}} \sqrt{1 - t^{2}} l n (t) - \frac{d t}{\sqrt{1 - t^{2}}}$ $= \int_{\frac{\sqrt{2}}{2}}^{1} l n (t) d t = {[t l n (t) - t_{}]}_{\frac{\sqrt{2}}{2}}^{1}$ $= - 1 - \frac{\sqrt{2}}{2} l n (\frac{\sqrt{2}}{2}) + \frac{\sqrt{2}}{2}$ $= \frac{\sqrt{2}}{2} - 1 - \frac{\sqrt{2}}{2} {\frac{1}{2} l n (2) - l n (2)}$ $= \frac{\sqrt{2}}{2} - 1 + \frac{1}{4} \sqrt{2} l n (2) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 15/May/18
	$t=cosx dt=−sinxdx ∫_1 ^(1/(√2)) −lnt dt I′=∫lnt dt =lnt×t−t so I=−∣tlnt−t∣_1 ^(1/(√2)) =−[{(1/(√2))ln((1/(√2)))−(1/(√2))}−{1.ln1−1}] =−[(1/(√2))(−(1/2)ln2)−(1/(√2))}−{−1}] =−[(−(1/(2(√2)))ln2−(1/(√2))}+1] =(1/(2(√2)))ln2+(1/(√2))−1 pls check$
	$t = c o s x d t = - s i n x d x$ $\int_{1}^{\frac{1}{\sqrt{2}}} - l n t d t$ $I^{'} = \int l n t d t$ $= l n t \times t - t$ $s o I = - ∣ t l n t - t ∣_{1}^{\frac{1}{\sqrt{2}}}$ $= - [{\frac{1}{\sqrt{2}} l n (\frac{1}{\sqrt{2}}) - \frac{1}{\sqrt{2}}} - {1 . l n 1 - 1}]$ $= - [\frac{1}{\sqrt{2}} (- \frac{1}{2} l n 2) - \frac{1}{\sqrt{2}}} - {- 1}]$ $= - [(- \frac{1}{2 \sqrt{2}} l n 2 - \frac{1}{\sqrt{2}}} + 1]$ $= \frac{1}{2 \sqrt{2}} l n 2 + \frac{1}{\sqrt{2}} - 1$ $p l s c h e c k$
	Commented by math khazana by abdo last updated on 15/May/18

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