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Question 89 of 104

Q.Evaluate: ∫0π/2sin⁡2x dx\displaystyle\int_0^{\pi/2} \sin^2 x\,\mathrm{d}x

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2019Subjective· 2mImportance★★★★★
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Use the half-angle identity sin⁡2x=1−cos⁡2x2\sin^2x=\dfrac{1-\cos2x}{2}.

∫0π/2sin⁡2x dx=∫0π/21−cos⁡2x2 dx=12[x−sin⁡2x2]0π/2\int_0^{\pi/2}\sin^2x\,dx = \int_0^{\pi/2}\dfrac{1-\cos2x}{2}\,dx = \dfrac12\left[x-\dfrac{\sin2x}{2}\right]_0^{\pi/2}

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