Question

find f.

$f(t)=\sec (t)(\sec (t)+\tan (t)),\quad -\frac{\pi}{2}

$f(t)=$

Explanation:

Step1: Expand the function

$$\begin{align*} f^{\prime}(t)&=\sec(t)(\sec(t)+\tan(t))\\ &=\sec^{2}(t)+\sec(t)\tan(t) \end{align*}$$

Step2: Integrate the function

We know that \(\int\sec^{2}(t)dt = \tan(t)+C_1\) and \(\int\sec(t)\tan(t)dt=\sec(t)+C_2\). So \(\int f^{\prime}(t)dt=\int(\sec^{2}(t)+\sec(t)\tan(t))dt=\tan(t)+\sec(t)+C\), where \(C = C_1 + C_2\). Then \(f(t)=\tan(t)+\sec(t)+C\)

Step3: Use the initial condition

Given \(f(\frac{\pi}{4})=-7\). Substitute \(t = \frac{\pi}{4}\) into \(f(t)\):

$$
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Solve for \(C\): \(C=-8 - \sqrt{2}\)

Answer:

\(f(t)=\tan(t)+\sec(t)-8-\sqrt{2}\)