MCQ
$\int_0^\pi {\frac{{x\tan x}}{{\sec x + \tan x}}} \,dx = $
- A$\frac{\pi }{2} - 1$
- B$\pi \left( {\frac{\pi }{2} + 1} \right)$
- C$\frac{\pi }{2} + 1$
- ✓$\pi \left( {\frac{\pi }{2} - 1} \right)$
==> $2I = \frac{\pi }{2}\int_0^\pi {\frac{{\tan x}}{{\sec x + \tan x}}dx = \frac{\pi }{2}\int_0^\pi {\frac{{\sin x}}{{1 + \sin x}}dx} } $
$=\frac{\pi }{2}\left[ {\int_0^\pi {1dx - \int_0^\pi {\frac{{dx}}{{1 + \sin x}}} } } \right]$
On solving, we get $I = \frac{{{\pi ^2}}}{2} - \pi = \pi \left( {\frac{\pi }{2} - 1} \right)$.
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$ \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(2+\lambda) \hat{\mathrm{i}}+(1-3 \lambda) \hat{\mathrm{j}}+(3+4 \lambda) \hat{\mathrm{k}}, \lambda \in \mathbb{R} $
$ \mathrm{L}_2: \overrightarrow{\mathrm{r}}=2(1+\mu) \hat{\mathrm{i}}+3(1+\mu) \hat{\mathrm{j}}+(5+\mu) \hat{k}, \mu \in \mathbb{R}$
is $\frac{\mathrm{m}}{\sqrt{\mathrm{n}}}$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then the value of $\mathrm{m}+\mathrm{n}$ equals.