- #1
Tony1
- 17
- 0
How may we go about to show that,
$$\int_{0}^{1}t\cos(2t\pi)\tan(t\pi)\ln[\sin(t\pi)]\mathrm dt=\color{green}{1\over \pi}\cdot\color{blue}{{\ln 2\over 2}(1-\ln 2)}$$
$$\int_{0}^{1}t\cos(2t\pi)\tan(t\pi)\ln[\sin(t\pi)]\mathrm dt=\color{green}{1\over \pi}\cdot\color{blue}{{\ln 2\over 2}(1-\ln 2)}$$