Question Number 84942 by M±th+et£s last updated on 17/Mar/20
$$\int_{\mathrm{0}} ^{{x}} {sinh}\left({x}−{t}\right)\:{cosh}\left({t}\right)\:{dt} \\ $$
Answered by mind is power last updated on 17/Mar/20
![sh(x)ch(y)=(1/2){sh(x+y)+sh(x−y)} =∫_0 ^x (1/2){sh(x)+sh(x−2t)dt =((sh(x)x)/2)+[−((ch(x−2t))/2)]_0 ^x =((xsh(x))/2)−((ch(−x))/2)+((ch(x))/2)=((xsh(x))/2)](https://www.tinkutara.com/question/Q84943.png)
$${sh}\left({x}\right){ch}\left({y}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left\{{sh}\left({x}+{y}\right)+{sh}\left({x}−{y}\right)\right\} \\ $$$$=\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{1}}{\mathrm{2}}\left\{{sh}\left({x}\right)+{sh}\left({x}−\mathrm{2}{t}\right){dt}\right. \\ $$$$=\frac{{sh}\left({x}\right){x}}{\mathrm{2}}+\left[−\frac{{ch}\left({x}−\mathrm{2}{t}\right)}{\mathrm{2}}\right]_{\mathrm{0}} ^{{x}} \\ $$$$=\frac{{xsh}\left({x}\right)}{\mathrm{2}}−\frac{{ch}\left(−{x}\right)}{\mathrm{2}}+\frac{{ch}\left({x}\right)}{\mathrm{2}}=\frac{{xsh}\left({x}\right)}{\mathrm{2}} \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 17/Mar/20
Commented by M±th+et£s last updated on 17/Mar/20
$${sir}\:{can}\:{you}\:{tell}\:{me}\:{where}\:{i}\:{make}\:{a}\:{fault} \\ $$
Commented by mind is power last updated on 17/Mar/20
$${lign}\:\mathrm{6}\:\:\frac{{e}^{{x}} }{−\mathrm{2}}−\frac{{e}^{−{x}} }{\mathrm{2}}=−{ch}\left({x}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left({xe}^{{x}} −{xe}^{−{x}} −\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{{x}} +{e}^{−{x}} \right)−\left(\frac{{e}^{{x}} }{−\mathrm{2}}−\frac{{e}^{−{x}} }{\mathrm{2}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{xsh}\left({x}\right)−{ch}\left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{{x}} +{e}^{−{x}} \right)\right) \\ $$$$=\frac{{xsh}\left({x}\right)}{\mathrm{2}}−\frac{{ch}\left({x}\right)}{\mathrm{4}}+\frac{{ch}\left({x}\right)}{\mathrm{4}}=\frac{{xsh}\left({x}\right)}{\mathrm{2}} \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 17/Mar/20
$${thank}\:{you}\:{so}\:{much}\:{sir}\: \\ $$