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Question: \(\int_{0}^{\pi/2}\frac{\cos\theta}{\sqrt{4–\sin^{2}\theta}}\) dq is equal to -...

∫0π/2cos⁡θ4–sin⁡2θ\int_{0}^{\pi/2}\frac{\cos\theta}{\sqrt{4–\sin^{2}\theta}} dq is equal to -

A

π2\frac{\pi}{2}

B

π6\frac{\pi}{6}

C

π3\frac{\pi}{3}

D

π5\frac{\pi}{5}

Answer

π6\frac{\pi}{6}

Explanation

Solution

sin q = t Ž cos q dq = dt

∫01dt/22–t2\int_{0}^{1}{dt/\sqrt{2^{2}–t^{2}}} = [sin–1(t/2)]01 Ž p/6