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0.025*(cos(3.7502*x)-cosh(3.7502*x)+(sin(3.7502*0.5)-sinh(3.

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0.025*(cos(3.7502*x)-cosh(3.7502*x)+(sin(3.7502*0.5)-sinh(3.7502*0.5))*(sin(3.7502*x)-sinh(3.7502*x))/(cos(3.7502*0.5)+cosh(3.7502*0.5)))^2
从0积分到0.5,
0.025*{cos(3.7502*x)-cosh(3.7502*x)+[sin(3.7502*0.5)-sinh(3.7502*0.5)]*[sin(3.7502*x)-sinh(3.7502*x)]/[cos(3.7502*0.5)+cosh(3.7502*0.5)]}^2
0.025*(cos(3.7502*x)-cosh(3.7502*x)+(sin(3.7502*0.5)-sinh(3.
clear
>> syms x
>> int=int(0.025*(cos(3.7502*x)-cosh(3.7502*x)+(sin(3.7502*0.5)-sinh(3.7502*0.5))*(sin(3.7502*x)-sinh(3.7502*x))/(cos(3.7502*0.5)+cosh(3.7502*0.5)))^2,0,0.5)
int =
(2697016455722038803766050922175625*sin(18751/5000))/3509583865900115839992672238798139644 - (9000972537865356001819891956864625*cosh(18751/10000)*sin(18751/10000))/877395966475028959998168059699534911 + (9000972537865356001819891956864625*cosh(18751/10000)*sinh(18751/10000))/1754791932950057919996336119399069822 + (1255384013175183500*sin(18751/10000)*sinh(18751/10000))/128265551756398210369 - (2697016455722038803766050922175625*sinh(18751/10000)*(2*cos(18751/20000)^2 - 1))/877395966475028959998168059699534911 + (627692006587591750*cos(18751/10000)^2)/128265551756398210369 - (627692006587591750*cosh(18751/10000)^2)/128265551756398210369 + 1/80
>> int1=double(int)
int1 =
0.0125
故最终的积分值为0.0125
0.025*(cos(3.7502*x)-cosh(3.7502*x)+(sin(3.7502*0.5)-sinh(3. 怎么用MATLAB解超越方程cos(x)-2*x*sin(x)-x-0.5*sin(x)*cosh(x)/sinh(x) matlab求定积分 c1(cosh(k*x)+cos(k*x))+c2(sinh(k*x)+sin(k*x))]^2 谁知道用matlab怎么求解1-cos(x)*cosh(x)-sin(x)*sinh(x)=0这个方程的前十个解么? sinh (x+y)=sinh x cosh y+cosh xsinh y 计算器上sinh、sin、sin^-1；cos、cosh、cos^-1；tanh、tan、tan^-1的区别 Free pascal中 sin(n);cos(n);tan(n);sinh(n);cosh(n);tanh(n); 怎 双曲函数证明有没有方法可以只用 e^x=cosh(x)+sinh(x)证明 cosh(2x)=cosh^2 (x)+si 谁有类似matlab那样可以实现sin,cos,tan,sinh,cosh,tanh,log,log10,exp和pow tan /asin /acos /atan /atan2(y,x) /sinh /cosh /tanh 这些都是函数,能 cos(x)*cosh(x)+1=0 solve('cos(t)*cosh(t)+1=0') ans = 1.87510 cosh (x + y)