Question

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

\(r\left( \theta \right) = sec\theta \,tan\theta - 2{e^\theta }\)

Short answer

The most general antiderivative of the function is \(R\left( \theta \right) = sec\theta \, - 2{e^\theta } + C\).

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

• \(cf\left( x \right) = cF\left( x \right)\)
• \(f\left( x \right) + g\left( x \right) = F\left( x \right) + G\left( x \right)\)
• \({e^x} = {e^x}\)
• \(secx\,\,tanx = secx\)

Finding the most general antiderivative of the function

Given the function is,

\(r\left( \theta \right) = sec\theta \,tan\theta - 2{e^\theta }\)

Using the Theorem 1 and the standard formula.

\(R\left( \theta \right) = sec\theta \, - 2{e^\theta } + C\)

The most general antiderivative of the function is \(R\left( \theta \right) = sec\theta \, - 2{e^\theta } + C\).

Checking by differentiation

\(R\left( \theta \right) = sec\theta \, - 2{e^\theta } + C\)

Differentiating with respect to \(\theta \).

\(\begin{aligned}{l}r\left( \theta \right) &= \frac{d}{{d\theta }}\left( {sec\theta \, - 2{e^\theta } + C} \right)\\r\left( \theta \right) = sec\theta \,tan\theta - 2{e^\theta } + 0\\r\left( \theta \right) &= sec\theta \,tan\theta - 2{e^\theta }\end{aligned}\)

Since the original function is obtained so antiderivative is correct.