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Question Number 32353 by abdo imad last updated on 23/Mar/18

$$calculate{\int }_0^{\frac{\pi }{4}}cos\left(x\right)ln\left(cos\left(x\right)\right)dx.$$

Answered by sma3l2996 last updated on 25/Mar/18

$$A={\int }_0^{\pi /4}cos\left(x\right)ln\left(cosx\right)dx$$$$byparts$$$$u=ln\left(cosx\right)\Rightarrow u^{\prime }=-tanx$$$$v^{\prime }=cosx\Rightarrow v=sinx$$$$A={\left[sin\left(x\right)ln\left(cosx\right)\right]}_0^{\pi /4}+{\int }_0^{\pi /4}\frac{{sin}^{2}x}{cosx}dx=-\frac{\sqrt{2}}{4}ln\left(2\right)+{\int }_0^{\pi /4}(\frac{1}{cosx}-cosx)dx$$$$lett=tan\left(x/2\right)$$$$A=-\frac{\sqrt{2}}{4}ln\left(2\right)-{\left[sinx\right]}_0^{\pi /4}+{\int }_0^{tan\left(\pi /8\right)}\frac{dt}{1-t^2}$$$$=-\frac{\sqrt{2}}{4}ln\left(2\right)-\frac{\sqrt{2}}{2}+\frac{1}{2}{[ln\mid \frac{1+t}{1-t}\mid ]}_0^{tan\left(\pi /8\right)}$$$$A=-\frac{\sqrt{2}}{4}ln\left(2\right)-\frac{\sqrt{2}}{2}+\frac{1}{2}ln(\frac{2}{1-tan\left(\pi /8\right)}-1)$$

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