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Question 92 of 104

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

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2020Subjective· 2mImportance★★★★★
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Use the identity cos⁡2x=1+cos⁡2x2\cos^2x = \dfrac{1+\cos2x}{2}.

∫0π/2cos⁡2x dx=∫0π/21+cos⁡2x2 dx=[x2+sin⁡2x4]0π/2\int_0^{\pi/2}\cos^2x\,dx = \int_0^{\pi/2}\dfrac{1+\cos2x}{2}\,dx = \left[\dfrac{x}{2}+\dfrac{\sin2x}{4}\right]_0^{\pi/2}

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