MHB 242t.08.02.41 Find the area of the region bounded by

[karush]

$\tiny{242t.08.02.41}$
$\textsf{Find the area of the region bounded above by}$
$\textsf{$y=8\cos{x}$ and below by $4\sec{x}$}$
$\textsf{and the limits are $-\frac{\pi}{4}\le x \le \frac{\pi}{4}$}$
\begin{align*} \displaystyle
I_{41}&=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (8\cos{x}-4\sec{x})\,dx
\end{align*}
$\textit{just want to see if this is set up right before step on the gas pedal}$

 

[MarkFL]

$\tiny{242t.08.02.41}$
$\textsf{Find the area of the region bounded above by}$
$\textsf{$y=8\cos{x}$ and below by $4\sec{x}$}$
$\textsf{and the limits are $-\frac{\pi}{4}\le x \le \frac{\pi}{4}$}$
\begin{align*}
I_{41}&=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (8\cos{x}-4\sec{x})\,dx
\end{align*}
$\textit{just want to see if this is set up right before step on the gas pedal}$

Yes, you have set this up correctly, however, I would take advantage of the symmetry of the integrand (even function) and write:

$$I=8\int_0^{\frac{\pi}{4}} 2\cos(x)-\sec(x)\,dx$$

 

[karush]

$\tiny{242t.08.01.41}$
\begin{align*} \displaystyle
I_{41}&=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (8\cos{x}-4\sec{x})\,dx\\
&=8\int_0^{\frac{\pi}{4}} 2\cos(x)-\sec(x)\,dx \\
&=16\int_0^{\frac{\pi}{4}} \cos(x) \,dx -8\int_0^{\frac{\pi}{4}} \sec(x)\,dx \\
&=\left[16\sin{x} -8 \ln{\cos{x}}\right]_{0}^{\frac{\pi} {4}}
\end{align*}

ok?

 

[MarkFL]

$\tiny{242t.08.01.41}$
\begin{align*} \displaystyle
I_{41}&=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (8\cos{x}-4\sec{x})\,dx\\
&=8\int_0^{\frac{\pi}{4}} 2\cos(x)-\sec(x)\,dx \\
&=16\int_0^{\frac{\pi}{4}} \cos(x) \,dx -8\int_0^{\frac{\pi}{4}} \sec(x)\,dx \\
&=\left[16\sin{x} -8 \ln{\cos{x}}\right]_{0}^{\frac{\pi} {4}}
\end{align*}

ok?

I do agree that:

$$\int \cos(x)\,dx=\sin(x)+C$$ since $$\frac{d}{dx}\left(\sin(x)+C\right)=\cos(x)$$

But, you stated:

$$\int \sec(x)\,dx=\ln|\cos(x)|+C$$ and $$\frac{d}{dx}\left(\ln|\cos(x)|+C\right)=\frac{-\sin(x)}{\cos(x)}=-\tan(x)\ne\sec(x)$$

What you want is:

$$\int \sec(x)\,dx=\ln|\sec(x)+\tan(x)|+C$$ since $$\frac{d}{dx}\left(\ln|\sec(x)+\tan(x)|+C\right)=\frac{\sec(x)\tan(x)+\sec^2(x)}{\sec(x)+\tan(x)}=\sec(x)$$ :D