Question

For $x>-1\text{and}f\left(x\right)={\int }_0^{\frac{\pi }{4}}{\log }_e(1+x\tan z)dz,$ then $f\left(\frac{1}{2}\right)-f\left(\frac{1}{3}\right)=$

• A. $0$
• B. $\left(\frac{\pi }{8}\right)lo{g}_e\left(\frac{3}{4}\right)$
• C. $\frac{\pi }{8}lo{g}_e\left(\frac{9}{8}\right)$
• D. $\frac{\pi }{4}lo{g}_e\left(\frac{4}{3}\right)$

Answer 1

The correct option is C $\frac{\pi }{8}lo{g}_e\left(\frac{9}{8}\right)$

$f\left(\frac{1}{2}\right)-f\left(\frac{1}{3}\right)={\int }_0^{\frac{\pi }{4}}{\log }_e\left(\right(\frac{2+\tan z}{3+\tan z}).\frac{3}{2})dz$

$I={\int }_0^{\frac{\pi }{4}}{\log }_e\left(\frac{2+\tan z}{3+\tan z}\right)dz+{\int }_0^{\frac{\pi }{4}}{\log }_e\left(\frac{3}{2}\right)dz$

$I={\int }_0^{\frac{\pi }{4}}{\log }_e\left(\frac{2+\tan (\frac{\pi }{4}-z)}{3+\tan (\frac{\pi }{4}-z)}\right)dz+\frac{\pi }{4}{\log }_e\left(\frac{3}{2}\right)$

$I={\int }_0^{\frac{\pi }{4}}{\log }_e\left(\frac{3+\tan z}{2+\tan z}\right)dz+{\int }_0^{\frac{\pi }{4}}{\log }_e\frac{1}{2}dz+\frac{\pi }{4}{\log }_e\left(\frac{3}{2}\right)$

$2I=0+\frac{\pi }{4}{\log }_e\left(\frac{9}{8}\right)$

$I=\frac{\pi }{8}{\log }_e\left(\frac{9}{8}\right)$