The correct option is D x+c We know that cosh x=e^x+e^ x2 and sinh x=e^x e^ x2 Hence, cosh x+sinh x =e^x+e^ x2+e^x e^ x2 =2(e^x) ...

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Circular Measurement of Angle

∫exdx/cosh x+...

Question

$\int \frac{e^{x} d x}{cosh x + sinh x} =$

log cosh x + c

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tan x + cot x + c

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\frac{1}{2} e^{2 x} + c

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x + c

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Solution

The correct option is D $x + c$
We know that
$cosh x = \frac{e^{x} + e^{- x}}{2}$ and
$sinh x = \frac{e^{x} - e^{- x}}{2}$
Hence, $cosh x + sinh x$
$= \frac{e^{x} + e^{- x}}{2} + \frac{e^{x} - e^{- x}}{2}$
$= \frac{2 (e^{x})}{2}$
$= e^{x}$ .
Hence, $\int \frac{e^{x}}{cosh x + sinh x} d x$
$= \int \frac{e^{x}}{e^{x}} d x$
$= \int d x$
$= x + c$

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