Question

Find the most general antiderivative of the function. (Check your answer by differentiation.)

\(f\left( t \right) = sint\, + 2\,sinht\)

Short answer

The most general antiderivative of the function is \(F\left( t \right) = - cost\, + 2\,cosht + C\).

Step-by-step solution

Theorem 1

If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is\(F\left( x \right) + C\), where C is an arbitrary constant.

Antidifferentiation formula

• \(cf\left( x \right) = cF\left( x \right)\)

• \(f\left( x \right) + g\left( x \right) = F\left( x \right) + G\left( x \right)\)

• \(sinx = - cosx\)

• \(sinhx = coshx\)

Finding the most general antiderivative of the function

Given the function is,

\(f\left( t \right) = sint\, + 2\,sinht\)

For finding antiderivative using the Theorem 1 and the standard formula

\(F\left( t \right) = - cost\, + 2\,cosht + C\)

The most general antiderivative of the function is \(F\left( t \right) = - cost\, + 2\,cosht + C\)

Checking by differentiation

\(F\left( t \right) = - cost\, + 2\,cosht + C\)

Differentiating with respect to t.

\(\begin{aligned}{l}f\left( t \right) = \frac{d}{{dt}}\left( { - cost\, + 2\,cosht + C} \right)\\f\left( t \right) = - \left( { - sint} \right)\, + 2\,sinht + 0\\f\left( t \right) = sint\, + 2\,sinht\end{aligned}\)

Since the original function is obtained so antiderivative is correct.