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Question Number 208280 by Shrodinger last updated on 10/Jun/24

$$L={\int }_0^{\frac{4}{\pi }}ln\left(cosx\right)dx$$

Answered by Berbere last updated on 10/Jun/24

$$={\int }_0^{\frac{\pi }{4}}\frac{ln(\frac{cos\left(x\right)}{sin\left(x\right)}.sin\left(x\right)cos\left(x\right))dx}{2}=\frac{1}{2}{\int }_0^{\frac{\pi }{4}}ln\left(cot\left(x\right)\right)dx+\frac{1}{2}{\int }_0^{\frac{\pi }{4}}ln\left(\frac{sin\left(2x\right)}{2}\right)dx$$$$tan\left(x\right)\to t;2x\to t$$$$=-\frac{1}{2}{\int }_0^1\frac{ln\left(t\right)}{1+t^2}dt+\frac{1}{4}{\int }_0^{\frac{\pi }{2}}ln\left(sin\left(y\right)\right)dy-\frac{ln\left(2\right)}{2}.\frac{\pi }{4}$$$$=-\frac{1}{2}{\int }_0^1\sum _{n⩾0}{(-1)}^{n}ln\left(t\right)t^{2n}dt+\frac{1}{4}.-\frac{\pi }{2}ln\left(2\right)-\frac{\pi ln\left(2\right)}{8}$$$$=-\frac{1}{2}\sum _{n⩾0}-\frac{1}{{(2n+1)}^2}-\frac{\pi ln\left(2\right)}{4};G=\frac{{(-1)}^n}{{(2n+1)}^2}=\beta (-1)Catalaneconstant$$$$=\frac{1}{2}G-\frac{\pi ln\left(2\right)}{4}$$

Commented by Shrodinger last updated on 11/Jun/24

$$thankssir..$$

Commented by Shrodinger last updated on 11/Jun/24

$$SirItis{\int }_0^{\frac{4}{\pi }}ln\left(cosx\right)dx$$

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