Calculus: Early Transcendentals, Chapter 4, 4.9, Section 4.9, Problem 18

int f(t) dt=int sin(t)+2 sinh(t) dt = int sin(t) dt + 2 int sinh(t) dt
We note that the derivative of cosh(t)=sinh(t) =>d/dx cosh(t) = sinh(t)=>d/dx sinh(t)=cosh(t)
This means that
int sinh(t) dt = cosh(t) +c_2 . We already know the anti-derivative of sin(t) , therefore the integral of f(t) is as follows:
int f(t) dt = -cos(t) + c_1 + 2 cosh(t) + c_2 = -cos(t) + cosh(t) + c , where c=c_1+c_2 is a constant. To check that this solution works we differentiate,
d/dx (-cos(t)+cosh(t)+c)=-(-sin(t))+sinh(t)+0=sin(t)+sinh(t) as we intended.

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