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GUJCET · Maths · Integrals

\(\int_0^{\pi / 4} \sqrt{1+\sin 2 x} d x=\) ___________.

  1. A 2
  2. B 1
  3. C \(\frac{1}{2}\)
  4. D 0
Verified Solution

Answer & Solution

Correct Answer

(B) 1

Step-by-step Solution

Detailed explanation

\(\int_0^{\pi / 4} \sqrt{1+\sin 2 x} d x = \int_0^{\pi / 4} \sqrt{(\sin x+\cos x)^2} d x\) \(= \int_0^{\pi / 4} (\sin x+\cos x) d x\)