Question

${\int }_0^{\frac{x}{2}}logtanxdx=$ [MP PET 1994; RPET 1995, 96, 97]

• A. $-\left(\frac{\pi }{2}\right)log2$
• B. $\pi log\frac{1}{2}$
• C. $-\pi log\frac{1}{2}$
• D. $\frac{\pi }{2}log2$

Answer 1

The correct option is A $-\left(\frac{\pi }{2}\right)log2$

${\int }_0^{\frac{x}{2}}logsinxdx={\int }_0^{\frac{x}{2}}logcosxdx$
$\Rightarrow 2I={\int }_0^{\frac{x}{2}}logsinxcosxdx={\int }_0^{\frac{x}{2}}logsin2xdx-{\int }_0^{\frac{x}{2}}log2dx$
$=\frac{1}{2}{\int }_0^{x}logsintdt-\frac{\pi }{2}log2,(Putting2x=t)$
$\Rightarrow 2I=I-\frac{\pi }{2}log2\Rightarrow I=\frac{-\pi }{2}log2,\{\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(t\right)dt\}$