2017年3月5日 · = 1/2 x sqrt(x^2 - 1) - 1/2 cosh^(-1) (x) + C When you say you want the "anti-derivative" of int sqrt(x^2-1) dx, I am assuming you mean you want the "anti-derivative" that is int sqrt(x^2-1) dx. Otherwise you are looking for: int ( int sqrt(x^2-1) dx) \\ dx. For int sqrt(x^2-1) \\ dx, there is a convenient hyperbolic relationship: cosh^2 z - sin^2 z = 1 If we let x …

2017年7月24日 · I got = tanh(x)ln(tanh(x/2)) - 2arctan(tanh(x/2)) + C, " "x > 0 DISCLAIMER: VERY LONG ANSWER! Well, I don't work with hyperbolics much, but I do know that: the derivative of tanh(x) is still sech^2(x). we still have cosh(x) = 1/(sech(x)). we however, have that -sinh^2(x) + cosh^2(x) = 1, so that tanh^2(x) + sech^2(x) = 1 and 1 + csch^2(x) = coth^2(x). So, we have: = int sech^2(x)ln (tanh(x/2 ...

2017年10月15日 · How do you show that #cosh^2 x - sinh^2 x = 1# ? Trigonometry. 1 Answer

2018年4月21日 · Use identities and axioms to change only one side until it is identical to the other side. Prove: sinh(2x)coth(x) = cosh(2x)+1 Substitute sinh(2x) = 2sinh(x)cosh(x) 2sinh(x)cosh(x)coth(x) = cosh(2x)+1 Substitute coth(x) = cosh(x)/sinh(x) 2sinh(x)cosh(x)cosh(x)/sinh(x) = cosh(2x)+1 sinh(x)/sinh(x) to 1: 2cosh(x)cosh(x) = cosh(2x)+1 Write as a square: 2cosh^2(x) = cosh(2x)+1 One identity for ...

2017年5月9日 · cosh x = 1+x^2/2+x^4/24+o(x^4) The MacLaurin series for e^x is: e^x = sum_(n=0)^oo x^n/(n!) Truncating the series at n=4 we obtain the Taylor polynomial of degree 4 ...

2017年4月12日 · The answer is =x/(|x|sqrt(x^2+1) We need (coshx)'=sinhx cosh^2x-sinh^2x=1 sinh^2x=cosh^2x-1 Let rewrite ...

2018年5月16日 · The identity #cosh^2(x)- sinh^2(x)=1# tells us that the denominator is -1: #int x / ( sinh(x) + cosh(x) ) dx = int (xsinh(x)-xcosh(x)) / -1 dx# Eliminate the denominator by changing signs in the numerator: #int x / ( sinh(x) + cosh(x) ) dx = int xcosh(x)-xsinh(x) dx# Separate into two integrals: #int x / ( sinh(x) + cosh(x) ) dx = int xcosh(x ...

2015年4月2日 · let y = cosh(lnx) => y = 1/2*(e^(lnx) - e^(-lnx)) = 1/2*(e^(lnx) + e^(lnx^-1)) = 1/2(x + x^-1) (dy)/(dx) = 1/2(1 + (-1)*x^-2) = 1/2((x^2 - 1)/x^2) = (x^2 - 1)/(2x^2)

2016年6月7日 · How do you find the derivative of #y = e^cosh(2x)#? Calculus Basic Differentiation Rules Chain Rule. 1 Answer

2016年9月2日 · How do you find the integral of #1/(cos x)^2#? Calculus Introduction to Integration Integrals of Trigonometric Functions. 1 Answer