find the indefinite integral of the following f(t)=6cos(3t)+5e^-10t
To find the indefinite integral of f(t) = 6cos(3t) + 5e^(-10t), we can integrate the two terms separately.
For the first term, 6cos(3t), we can use the power rule for integrating trigonometric functions. The integral of cos(x) is sin(x), and since we have a coefficient of 6 and an argument of 3t, the integral becomes:
∫ 6cos(3t) dt = (6/3)sin(3t) + C = 2sin(3t) + C1,
where C is the constant of integration and C1 is a new constant representing the combination of the original constant of integration and the multiplied constant.
For the second term, 5e^(-10t), we can use the power rule for integrating exponential functions. The integral of e^(kx) is (1/k)e^(kx), so the integral becomes:
∫ 5e^(-10t) dt = (5/-10)e^(-10t) + C = -(1/2)e^(-10t) + C2,
where C2 is another constant of integration.
Now, putting the two integrals together, we have:
∫ f(t) dt = 2sin(3t) - (1/2)e^(-10t) + C,
where C is the combined constant of integration.
So, the indefinite integral of f(t) = 6cos(3t) + 5e^(-10t) is 2sin(3t) - (1/2)e^(-10t) + C.