Question

6 mark for review which of the following is an antiderivative of (f(x)=\tan(e^{x}+\frac{pi}{2})) on (0leq xleqln(\frac{pi}{2}))? a (sec^{2}(e^{x}+\frac{pi}{2})) b (\frac{ln|sec(e^{x}+\frac{pi}{2})|}{e^{x}}) c (int_{0}^{e^{x}}\tan(t + \frac{pi}{2})dt) d (int_{0}^{x}\tan(e^{t}+\frac{pi}{2})dt)

Explanation:

Step1: Recall the fundamental theorem of calculus

If $F(x)=\int_{a}^{x}f(t)dt$, then $F^\prime(x) = f(x)$. Here, we want an antiderivative of $y = f(x)=\tan(e^{x}+\frac{\pi}{2})$.

Step2: Analyze each option

• Option A: The derivative of $\sec^{2}(e^{x}+\frac{\pi}{2})$ is not $\tan(e^{x}+\frac{\pi}{2})$ by the chain - rule. The derivative of $\sec^{2}(u)$ with $u = e^{x}+\frac{\pi}{2}$ is $2\sec(u)\sec(u)\tan(u)e^{x}=2e^{x}\sec^{2}(e^{x}+\frac{\pi}{2})\tan(e^{x}+\frac{\pi}{2})$.
• Option B: Using the quotient - rule and the derivative of $\ln|\sec(u)|=\tan(u)$ (where $u = e^{x}+\frac{\pi}{2}$), the derivative of $\frac{\ln|\sec(e^{x}+\frac{\pi}{2})|}{e^{x}}$ is not $\tan(e^{x}+\frac{\pi}{2})$.
• Option C: By the chain - rule and the fundamental theorem of calculus, if $G(x)=\int_{0}^{e^{x}}\tan(t + \frac{\pi}{2})dt$, let $u = e^{x}$, then $G(x)=\int_{0}^{u}\tan(t+\frac{\pi}{2})dt$. By the chain - rule and the fundamental theorem of calculus, $G^\prime(x)=\tan(e^{x}+\frac{\pi}{2})\cdot e^{x}$, not $\tan(e^{x}+\frac{\pi}{2})$.
• Option D: According to the fundamental theorem of calculus, if $F(x)=\int_{0}^{x}\tan(e^{t}+\frac{\pi}{2})dt$, then $F^\prime(x)=\tan(e^{x}+\frac{\pi}{2})$.

Answer:

D. $\int_{0}^{x}\tan(e^{t}+\frac{\pi}{2})dt$