MCQ
$\int_{}^{} {\frac{{dx}}{{{e^x} + {e^{ - x}}}} = } $
- A${\tan ^{ - 1}}({e^{ - x}})$
- ✓${\tan ^{ - 1}}({e^x})$
- C$\log ({e^x} - {e^{ - x}})$
- D$\log ({e^x} + {e^{ - x}})$
$ = {\tan ^{ - 1}}({e^x}) + c$,$\{$Putting ${e^x} = t \Rightarrow {e^x}dx = dt.$$\}$
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$x \ne m\pi \pm \frac{\pi }{6};m \in Z$ and $f\left( {m\pi \pm \frac{\pi }{6}} \right) = 0$ , then
If $I_1 = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} \, f (\tan\, \theta + \cot\, \theta )\cdot sec^2\, \theta\, d\, \theta$ &
$I_2 = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} \, f (\tan\, \theta + \cot\, \theta )\cdot cosec^2\, \theta\, d \, \theta$ ,
then the ratio $\frac{{{I_1}}}{{{I_2}}}$ :