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Question Number 182271 by SANOGO last updated on 06/Dec/22

Answered by SEKRET last updated on 06/Dec/22

$$\frac{-\mathit{\pi }}{2}\cdot \mathbf{ln}\left(2\right)$$

Answered by SEKRET last updated on 06/Dec/22

$${\int }_{\mathbf{a}}^{\mathbf{b}}\mathbf{f}\left(\mathbf{x}\right)\mathbf{dx}={\int }_{\mathbf{a}}^{\mathbf{b}}\mathbf{f}(\mathbf{a}+\mathbf{b}-\mathbf{x})\mathbf{dx}$$$${\int }_0^{\frac{\mathit{\pi }}{2}}\mathbf{ln}\left(\mathbf{sinx}\right)\mathbf{dx}=\mathbf{I}$$$${\int }_0^{\frac{\mathit{\pi }}{2}}\mathbf{ln}\left(\mathbf{cos}\left(\mathbf{x}\right)\right)\mathbf{dx}=\mathbf{I}$$$${\int }_0^{\frac{\mathit{\pi }}{2}}\mathbf{ln}\left(\frac{\mathbf{sin}\left(2\mathbf{x}\right)}{2}\right)\mathbf{dx}=2\mathbf{I}$$$${\int }_0^{\frac{\mathit{\pi }}{2}}\mathbf{ln}\left(\mathbf{sin}\left(2\mathbf{x}\right)\right)\mathbf{dx}-{\int }_0^{\frac{\mathit{\pi }}{2}}\mathbf{ln}\left(2\right)\mathbf{dx}=2\mathbf{I}$$$${\int }_0^{\mathit{\pi }}\frac{1}{2}\cdot \mathbf{ln}\left(\mathbf{sin}\left(\mathbf{x}\right)\right)\mathbf{dx}-\frac{\mathit{\pi }}{2}\cdot \mathbf{ln}\left(2\right)=2\mathbf{I}$$$$\frac{1}{2}\cdot ({\int }_0^{\frac{\mathit{\pi }}{2}}\mathbf{ln}\left(\mathbf{sinx}\right)\mathbf{dx}+{\int }_{\frac{\mathit{\pi }}{2}}^{\mathit{\pi }}\mathbf{ln}\left(\mathbf{sinx}\right)\mathbf{dx})-\frac{\mathit{\pi }\cdot \mathbf{ln}\left(2\right)}{2}=2\mathbf{I}$$$$\frac{1}{2}(\mathbf{I}+\mathbf{I})-\frac{\mathit{\pi }}{2}\cdot \mathbf{ln}\left(2\right)=2\mathbf{I}$$$$\mathbf{I}=\frac{-\pi }{2}\cdot \mathbf{ln}\left(2\right)$$$$\mathbf{ABDULAZIZ}\mathbf{ABDUVALIYEV}$$

Commented by SEKRET last updated on 06/Dec/22

$${\int }_0^{\frac{\mathit{\pi }}{2}}\mathbf{ln}\left(\mathbf{sinx}\right)\mathbf{dx}={\int }_{\frac{\mathit{\pi }}{2}}^{{}^{\mathit{\pi }}}\mathbf{ln}\left(\mathbf{sinx}\right)\mathbf{dx}=\mathbf{I}$$

Commented by SANOGO last updated on 06/Dec/22

$$thankyou$$

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