Question Number 35676 by abdo imad last updated on 21/May/18
$${find}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:{ch}^{\mathrm{4}} {t}\:{dt} \\ $$
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)](https://www.tinkutara.com/question/Q35841.png)
$${we}\:{have}\:{f}\left({x}\right)=\:\int_{\mathrm{0}} ^{{x}} \:\:\left\{\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\right\}^{\mathrm{2}} {dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{{x}} \:\:\:\left\{\mathrm{1}\:+\mathrm{2}{ch}\left(\mathrm{2}{t}\right)\:+{ch}^{\mathrm{2}} \left(\mathrm{2}{t}\right)\right\}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{{x}} \:\left\{\:\:\mathrm{1}+\mathrm{2}{ch}\left(\mathrm{2}{t}\right)\:+\frac{\mathrm{1}+{ch}\left(\mathrm{4}{t}\right)}{\mathrm{2}}\right\}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{{x}} \:\left\{\:\mathrm{2}\:+\mathrm{4}{ch}\left(\mathrm{2}{t}\right)\:+\mathrm{1}\:+{ch}\left(\mathrm{4}{t}\right)\right\}{dt} \\ $$$$=\frac{\mathrm{3}{x}}{\mathrm{8}}\:+\:\left[\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{{x}} \:\:+\left[\:\frac{\mathrm{1}}{\mathrm{32}}{sh}\left(\mathrm{4}{t}\right)\right]_{\mathrm{0}} ^{{x}} \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{3}{x}}{\mathrm{8}}\:\:+\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{2}{x}\right)\:\:+\frac{\mathrm{1}}{\mathrm{32}}{sh}\left(\mathrm{4}{x}\right)\: \\ $$