Question Number 32353 by abdo imad last updated on 23/Mar/18
$${calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left({x}\right){ln}\left({cos}\left({x}\right)\right){dx}\:. \\ $$
Answered by sma3l2996 last updated on 25/Mar/18
![A=∫_0 ^(π/4) cos(x)ln(cosx)dx by parts u=ln(cosx)⇒u′=−tanx v′=cosx⇒v=sinx A=[sin(x)ln(cosx)]_0 ^(π/4) +∫_0 ^(π/4) ((sin^2 x)/(cosx))dx=−((√2)/4)ln(2)+∫_0 ^(π/4) ((1/(cosx))−cosx)dx let t=tan(x/2) A=−((√2)/4)ln(2)−[sinx]_0 ^(π/4) +∫_0 ^(tan(π/8)) (dt/(1−t^2 )) =−((√2)/4)ln(2)−((√2)/2)+(1/2)[ln∣((1+t)/(1−t))∣]_0 ^(tan(π/8)) A=−((√2)/4)ln(2)−((√2)/2)+(1/2)ln((2/(1−tan(π/8)))−1)](https://www.tinkutara.com/question/Q32465.png)
$${A}=\int_{\mathrm{0}} ^{\pi/\mathrm{4}} {cos}\left({x}\right){ln}\left({cosx}\right){dx} \\ $$$${by}\:{parts}\: \\ $$$${u}={ln}\left({cosx}\right)\Rightarrow{u}'=−{tanx} \\ $$$${v}'={cosx}\Rightarrow{v}={sinx} \\ $$$${A}=\left[{sin}\left({x}\right){ln}\left({cosx}\right)\right]_{\mathrm{0}} ^{\pi/\mathrm{4}} +\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \frac{{sin}^{\mathrm{2}} {x}}{{cosx}}{dx}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}{ln}\left(\mathrm{2}\right)+\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \left(\frac{\mathrm{1}}{{cosx}}−{cosx}\right){dx}\: \\ $$$${let}\:\:{t}={tan}\left({x}/\mathrm{2}\right) \\ $$$${A}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−\left[{sinx}\right]_{\mathrm{0}} ^{\pi/\mathrm{4}} +\int_{\mathrm{0}} ^{{tan}\left(\pi/\mathrm{8}\right)} \frac{{dt}}{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$=−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid\right]_{\mathrm{0}} ^{{tan}\left(\pi/\mathrm{8}\right)} \\ $$$${A}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}}{\mathrm{1}−{tan}\left(\pi/\mathrm{8}\right)}−\mathrm{1}\right) \\ $$