Question

The value of \[ I=\int_0^{\pi/2}\frac{(\sin x+\cos x)^2}{\sqrt{1+\sin 2x}}\,dx \] is:

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

When integrals contain \((\sin x+\cos x)^2\) and \(\sqrt{1+\sin 2x}\), try using the identity \((\sin x+\cos x)^2=1+\sin 2x\). This often cancels and simplifies beautifully.

Answer 1

The Correct Option is C

Step 1: Simplify numerator and denominator.
\[ (\sin x+\cos x)^2 = \sin^2x+\cos^2x+2\sin x\cos x = 1+\sin 2x. \] Thus, the integrand becomes: \[ \frac{1+\sin 2x}{\sqrt{1+\sin 2x}}=\sqrt{1+\sin 2x}. \]

Step 2: Simplify the expression inside the root.
\[ 1+\sin 2x = 1+2\sin x\cos x. \] Recall identity: \[ 1+\sin 2x = (\sin x+\cos x)^2. \] So, \[ \sqrt{1+\sin 2x}=\sin x+\cos x \text{(since both are nonnegative on } [0,\tfrac{\pi}{2}]). \]

Step 3: Evaluate the integral.
\[ I=\int_0^{\pi/2}(\sin x+\cos x)\,dx. \] Separate: \[ I=\int_0^{\pi/2}\sin x\,dx+\int_0^{\pi/2}\cos x\,dx. \] \[ =\Big[-\cos x\Big]_0^{\pi/2}+\Big[\sin x\Big]_0^{\pi/2}. \] \[ =(-\cos(\tfrac{\pi}{2})+\cos 0)+(\sin(\tfrac{\pi}{2})-\sin 0). \] \[ =(0+1)+(1-0)=2. \]

Final Answer:
\[ \boxed{2} \]