Question

If \(\int \frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\sqrt{8-\sin x}}dx=2\sin ^{-1}\left(\frac{\phi (x)}{3}\right)+c\), then \([\phi (x)]^{2}\) is equal to _____

• A. \(1+\cos x\)
• B. \(1+\sin x\)
• C. \(1-\sin x\)
• D. \(1-\cos x\)

Hint:

Substitute t = cos(x/2) + sin(x/2) so that 8 - sin x becomes 9 - t squared.

Answer 1

The Correct Option is B

Step 1: Differentiate the Answer Form
If the integral is $2\sin^{-1}\dfrac{\phi}{3}$, its derivative is $\dfrac{2\phi'}{\sqrt{9-\phi^{2}}}$.

Step 2: Match with the Integrand
The integrand has $\sqrt{8-\sin x}$ below, so $9-\phi^{2}=8-\sin x$, giving $\phi^{2}=1+\sin x$.

Step 3: Check the Numerator
With $\phi=\cos\dfrac{x}{2}+\sin\dfrac{x}{2}$, $2\phi'=\cos\dfrac{x}{2}-\sin\dfrac{x}{2}$, which matches the given numerator.

Final Answer:
Hence phi squared is 1 plus sin x. \[ \boxed{\text{(B) }1+\sin x} \]