Question

If $f'(x) = \tan^{-1}(\sec x + \tan x)$, $-\frac{\pi}{2}<x<\frac{\pi}{2}$ and $f(0) = 0$, then $f(1)$ is

• A. $\frac{\pi+1}{4}$
• B. $\frac{\pi+2}{4}$
• C. $\pi+\frac{1}{4}$
• D. $\frac{\pi-1}{4}$

Hint:

Trigonometry Tip: The reduction sequence $\sec x + \tan x \rightarrow \frac{1+\sin x}{\cos x} \rightarrow \tan(\frac{\pi}{4} + \frac{x}{2})$ is an extremely common pattern in calculus and should be memorized to save time.

Answer 1

The Correct Option is A

Concept: Calculus - Indefinite Integration, Inverse Trigonometric Simplification, and Initial Value Problems.

Step 1: Simplify the expression inside the inverse tangent function. Let's convert $\sec x$ and $\tan x$ to sines and cosines: $\sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1+\sin x}{\cos x}$.

Step 2: Apply half-angle formulas to convert to a tangent expression. We can write $1 = \cos^{2}(\frac{x}{2}) + \sin^{2}(\frac{x}{2})$ and $\sin x = 2\sin(\frac{x}{2})\cos(\frac{x}{2})$. This makes the numerator a perfect square: $(\cos\frac{x}{2} + \sin\frac{x}{2})^{2}$. Using the double angle formula for the denominator: $\cos x = \cos^{2}(\frac{x}{2}) - \sin^{2}(\frac{x}{2})$. The fraction becomes: $\frac{(\cos\frac{x}{2} + \sin\frac{x}{2})^{2}}{\cos^{2}\frac{x}{2} - \sin^{2}\frac{x}{2}} = \frac{(\cos\frac{x}{2} + \sin\frac{x}{2})^{2}}{(\cos\frac{x}{2} - \sin\frac{x}{2})(\cos\frac{x}{2} + \sin\frac{x}{2})} = \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}$.

Step 3: Transform into the standard tangent addition formula. Divide the numerator and denominator by $\cos(\frac{x}{2})$: $\frac{1 + \tan(\frac{x}{2})}{1 - \tan(\frac{x}{2})}$. This matches the identity for $\tan(\frac{\pi}{4} + \frac{x}{2})$.

Step 4: Substitute back into $f^{\prime}(x)$ and integrate. Now, $f^{\prime}(x) = \tan^{-1}\left(\tan(\frac{\pi}{4} + \frac{x}{2})\right) = \frac{\pi}{4} + \frac{x}{2}$. We integrate this to find $f(x)$: $f(x) = \int (\frac{\pi}{4} + \frac{x}{2})dx = \frac{\pi x}{4} + \frac{x^2}{4} + c$.

Step 5: Apply the initial condition to find $c$ and calculate $f(1)$. We are given $f(0) = 0$. Substituting $x=0$ yields $0 = 0 + 0 + c \implies c = 0$. Thus, the exact function is $f(x) = \frac{\pi x}{4} + \frac{x^2}{4}$. To find $f(1)$, simply substitute $x=1$: $f(1) = \frac{\pi(1)}{4} + \frac{(1)^2}{4} = \frac{\pi}{4} + \frac{1}{4} = \frac{\pi+1}{4}$. $$ \therefore \text{The value of } f(1) \text{ is } \frac{\pi+1}{4}. $$