Question

If \(\int \sqrt{1+\sec x}dx=2\sin ^{-1}(f(x))+c\), then

• A. \(f(x)=\sqrt{2}\sin \left(\frac{x}{2}\right)\)
• B. \(f(x)=\sqrt{2}\cos \left(\frac{x}{2}\right)\)
• C. \(f(x)=\frac{1}{\sqrt{2}}\sin \left(\frac{x}{2}\right)\)
• D. \(f(x)=\frac{1}{\sqrt{2}}\cos \left(\frac{x}{2}\right)\)

Hint:

Differentiate the right side and compare it with the integrand after rewriting 1 + sec x using half-angle formulas.

Answer 1

The Correct Option is A

Step 1: Differentiate Option (A)
For $f=\sqrt{2}\sin(x/2)$: $\dfrac{d}{dx}2\sin^{-1}f=\dfrac{2f'}{\sqrt{1-f^{2}}}$ with $f'=\dfrac{\sqrt{2}}{2}\cos\dfrac{x}{2}$.

Step 2: Simplify
$1-f^{2}=1-2\sin^{2}\dfrac{x}{2}=\cos x$, so the derivative is $\dfrac{\sqrt{2}\cos(x/2)}{\sqrt{\cos x}}$.

Step 3: Compare
This equals $\sqrt{\dfrac{2\cos^{2}(x/2)}{\cos x}}=\sqrt{1+\sec x}$, so option (A) works.

Final Answer:
Hence $f(x)=\sqrt{2}\sin\dfrac{x}{2}$. \[ \boxed{\text{(A) }f(x)=\sqrt{2}\sin \left(\dfrac{x}{2}\right)} \]