Key Concepts and Formulas

• Trigonometric Identities:
  • $\sec x = \frac{1}{\cos x}$
  • $\tan x = \frac{\sin x}{\cos x}$
  • $1 + \sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$
  • $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$
  • $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
• Integration: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$
• $\tan(\frac{\pi}{4}) = 1$

Step-by-Step Solution

Step 1: Rewrite $\sec x + \tan x$ in terms of $\sin x$ and $\cos x$. We are given $f'(x) = \tan^{-1}(\sec x + \tan x)$. We first express $\sec x$ and $\tan x$ in terms of $\sin x$ and $\cos x$: $f'(x) = \tan^{-1}\left(\frac{1}{\cos x} + \frac{\sin x}{\cos x}\right) = \tan^{-1}\left(\frac{1 + \sin x}{\cos x}\right)$

Step 2: Simplify the expression inside the $\tan^{-1}$ function. We can rewrite $1 + \sin x$ and $\cos x$ using half-angle identities. Recall that $1 = \sin^2(x/2) + \cos^2(x/2)$ and $\sin x = 2\sin(x/2)\cos(x/2)$. Then, $1 + \sin x = \sin^2(x/2) + \cos^2(x/2) + 2\sin(x/2)\cos(x/2) = (\cos(x/2) + \sin(x/2))^2$. Also, $\cos x = \cos^2(x/2) - \sin^2(x/2)$. Thus,

$f'(x) = \tan^{-1}\left(\frac{(\cos(x/2) + \sin(x/2))^2}{\cos^2(x/2) - \sin^2(x/2)}\right) = \tan^{-1}\left(\frac{(\cos(x/2) + \sin(x/2))^2}{(\cos(x/2) - \sin(x/2))(\cos(x/2) + \sin(x/2))}\right)$

$f'(x) = \tan^{-1}\left(\frac{\cos(x/2) + \sin(x/2)}{\cos(x/2) - \sin(x/2)}\right)$

Now, divide both numerator and denominator by $\cos(x/2)$: $f'(x) = \tan^{-1}\left(\frac{1 + \tan(x/2)}{1 - \tan(x/2)}\right)$

Step 3: Use the tangent addition formula. Recall that $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$. Let $A = \frac{\pi}{4}$ and $B = \frac{x}{2}$. Then $\tan(\frac{\pi}{4}) = 1$, so $f'(x) = \tan^{-1}\left(\frac{\tan(\pi/4) + \tan(x/2)}{1 - \tan(\pi/4)\tan(x/2)}\right) = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right)$

Step 4: Simplify using the given range of $x$. Since $-\frac{\pi}{2} < x < \frac{\pi}{2}$, we have $-\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{4}$. Therefore, $0 < \frac{\pi}{4} + \frac{x}{2} < \frac{\pi}{2}$. This means that $\frac{\pi}{4} + \frac{x}{2}$ lies in the range of the principal value of the inverse tangent function. Thus, $f'(x) = \frac{\pi}{4} + \frac{x}{2}$

Step 5: Integrate $f'(x)$ to find $f(x)$. Integrating both sides with respect to $x$, we get: $f(x) = \int \left(\frac{\pi}{4} + \frac{x}{2}\right) dx = \frac{\pi}{4}x + \frac{x^2}{4} + C$ where $C$ is the constant of integration.

Step 6: Use the initial condition $f(0) = 0$ to find $C$. We are given that $f(0) = 0$. Substituting $x=0$ into the expression for $f(x)$, we have: $f(0) = \frac{\pi}{4}(0) + \frac{(0)^2}{4} + C = 0$ Thus, $C = 0$. Therefore, $f(x) = \frac{\pi}{4}x + \frac{x^2}{4}$

Step 7: Calculate $f(1)$. We want to find $f(1)$. Substituting $x=1$ into the expression for $f(x)$, we get: $f(1) = \frac{\pi}{4}(1) + \frac{(1)^2}{4} = \frac{\pi}{4} + \frac{1}{4} = \frac{\pi + 1}{4}$

Common Mistakes & Tips

• Range of inverse tangent: Be mindful of the range of the inverse tangent function when simplifying $\tan^{-1}(\tan(u))$. In this case, the given range of $x$ ensures that $\frac{\pi}{4} + \frac{x}{2}$ lies within the principal range.
• Trigonometric identities: Remember the half-angle formulas and tangent addition formula to simplify the expression.
• Constant of integration: Don't forget to include the constant of integration when finding the indefinite integral. Use the initial condition to find the value of the constant.

Summary

We simplified the expression for $f'(x)$ using trigonometric identities and the tangent addition formula. We then integrated $f'(x)$ to find $f(x)$, and used the initial condition $f(0) = 0$ to find the constant of integration. Finally, we calculated $f(1)$ to obtain the desired result.

The final answer is $\boxed{\frac{\pi + 1}{4}}$, which corresponds to option (C).