Chapter 4.8, Problem 27E

To determine

To calculate: The function $f\left(x\right)$ from the equation $f^{\text{'}}\left(t\right)=2\cos t+{\sec }^{2}t$ with condition $f\left(\frac{\pi }{3}\right)=4,\frac{-\pi }{2}<t<\frac{\pi }{2}$ .

Answer to Problem 27E

The function $f\left(x\right)=2\sin t+{\tan }^{-1}t+4-2\sqrt{3}$ from the equation $f^{\text{'}}\left(t\right)=2\cos t+{\sec }^{2}t$ .

Explanation of Solution

Given information:

The equation is given as:

  $f^{\text{'}}\left(t\right)=2\cos t+{\sec }^{2}t$ and condition $f\left(\frac{\pi }{3}\right)=4,\frac{-\pi }{2}<t<\frac{\pi }{2}$ .

Formula used:

The function F is antiderivative of $f\left(x\right)$ on an interval $I$ if $F^{`}\left(x\right)=f\left(x\right)$ for all $x$ in $I$

The function $F$ is known as antiderivative of $f\left(x\right)$ on an interval $I$ the most general antiderivative of the function on interval $I$ is $F\left(x\right)+C$ , where $C$ is an arbitrary constant.

Calculation:

Consider the function,

  $f^{\text{'}}\left(t\right)=2\cos t+{\sec }^{2}t$

The antiderivative of $f\left(x\right)$ is,

  $F^{`}\left(x\right)=f\left(x\right)$

So, by reverse power rule Integrating both sides of equation,

  $\begin{array}{c}{\displaystyle \int F^{`}\left(x\right)}={\displaystyle \int f\left(x\right)}\\ F\left(x\right)={\displaystyle \int f\left(x\right)}\end{array}$

Therefore, integrating the function

  $\begin{array}{c}{\displaystyle \int f^{\text{'}}\left(t\right)}dt={\displaystyle \int 2\cos t+{\sec }^{2}tdt}\\ f\left(t\right)=2\sin t+{\tan }^{-1}t+C\end{array}$

The value of $t$ is alwaysbetween $\frac{-\pi }{2}<t<\frac{\pi }{2}$ because if put $t=\frac{\pi }{2},t=\frac{-\pi }{2}$ then equation is undefined so value of $\frac{-\pi }{2}<t<\frac{\pi }{2}$

Apply, the condition $f\left(\frac{\pi }{3}\right)=4,\frac{-\pi }{2}<x<\frac{\pi }{2}$ on equation for obtain $C$ ,

  $\begin{array}{c}f\left(\frac{\pi }{3}\right)=2\sin \left(\frac{\pi }{3}\right)+{\tan }^{-1}\left(\frac{\pi }{3}\right)+C\\ 4=2\frac{\sqrt{3}}{2}+\sqrt{3}+C\\ 4-2\sqrt{3}=C\end{array}$

Put the value of $C$ in the equation,

  $f\left(x\right)=2\sin t+{\tan }^{-1}t+4-2\sqrt{3}$

Thus, the particular solution is $f\left(x\right)=2\sin t+{\tan }^{-1}t+4-2\sqrt{3}$ .