find f(x)=∫_0 ^x ch^4 t dt


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Question Number 35676 by abdo imad last updated on 21/May/18

$f i n d f (x) = \int_{0}^{x} {c h}^{4} t d t$
Commented by abdo mathsup 649 cc last updated on 24/May/18
$we have f(x)= ∫_0 ^x {((1+ch(2t))/2)}^2 dt = (1/4) ∫_0 ^x {1 +2ch(2t) +ch^2 (2t)}dt =(1/4) ∫_0 ^x { 1+2ch(2t) +((1+ch(4t))/2)}dt =(1/8) ∫_0 ^x { 2 +4ch(2t) +1 +ch(4t)}dt =((3x)/8) + [(1/4)sh(2t)]_0 ^x +[ (1/(32))sh(4t)]_0 ^x f(x)= ((3x)/8) +(1/4)sh(2x) +(1/(32))sh(4x)$
$w e h a v e f (x) = \int_{0}^{x} {\frac{1 + c h (2 t)}{2}}^{2} d t$ $= \frac{1}{4} \int_{0}^{x} {1 + 2 c h (2 t) + {c h}^{2} (2 t)} d t$ $= \frac{1}{4} \int_{0}^{x} {1 + 2 c h (2 t) + \frac{1 + c h (4 t)}{2}} d t$ $= \frac{1}{8} \int_{0}^{x} {2 + 4 c h (2 t) + 1 + c h (4 t)} d t$ $= \frac{3 x}{8} + {[\frac{1}{4} s h (2 t)]}_{0}^{x} + {[\frac{1}{32} s h (4 t)]}_{0}^{x}$ $f (x) = \frac{3 x}{8} + \frac{1}{4} s h (2 x) + \frac{1}{32} s h (4 x)$
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