The correct option is C 1/2tan^ 1( sinhx/2)+c #160;∫cos hxcos h^2x+3dx =∫cos hx4+sinh^2xdx sinh x= t cosh^2x dx= dt #160;∫dt4+t^2= ∫dt ...

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Limit

∫cosh x/cosh2...

Question

$\int \frac{cosh x}{{cosh}^{2} x + 3} d x =$

{tan}^{- 1} (\frac{s i n h x}{2}) + c

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\frac{1}{2} {tan}^{- 1} (\frac{s i n h x}{2}) + c

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\frac{1}{2} {tan}^{- 1} (sinh x) + c

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{tan}^{- 1} (sinh x) + c

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Solution

The correct option is C $\frac{1}{2} {tan}^{- 1} (\frac{s i n h x}{2}) + c$
$\int \frac{c o s h x}{c o s h^{2} x + 3} d x = \int \frac{c o s h x}{4 + s i n h^{2} x} d x$
$sinh x = t$
${cosh}^{2} x d x = d t$
$\int \frac{d t}{4 + t^{2}} = \int \frac{d t}{4 + t^{2}}$
$= \frac{1}{2} t a n^{- 1} (^{t} /_{2}) + c$
$= \frac{1}{2} t a n^{- 1} (\frac{s i n h x}{2}) + c$
$\int \frac{c o s h x}{c o s h^{2} x + 3} d x = \frac{1}{2} t a n^{- 1} (\frac{s i n h x}{2}) + c$
Hence, option 'B' is correct.

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