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Find-0-1-4-sin-2-x-2-cos-2-x-2-dx-




Question Number 201462 by hardmath last updated on 06/Dec/23
Find:  ∫_0 ^( 𝛑)  (√(1 - 4 sin^2  (x/2) cos^2  (x/2))) dx = ?
$$\mathrm{Find}: \\ $$$$\int_{\mathrm{0}} ^{\:\boldsymbol{\pi}} \:\sqrt{\mathrm{1}\:-\:\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\frac{\mathrm{x}}{\mathrm{2}}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\mathrm{x}}{\mathrm{2}}}\:\mathrm{dx}\:=\:? \\ $$
Answered by esmaeil last updated on 06/Dec/23
=∫_0 ^π (√((1−(2sin(x/2)cos(x/2))^2 ))dx=  ∫_0 ^π ∣cosx∣dx=∫^(π/2) _0 cosxdx−∫_(π/2) ^π cosxdx=2
$$=\int_{\mathrm{0}} ^{\pi} \sqrt{\left(\mathrm{1}−\left(\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} \right.}{dx}= \\ $$$$\int_{\mathrm{0}} ^{\pi} \mid{cosx}\mid{dx}=\underset{\mathrm{0}} {\int}^{\frac{\pi}{\mathrm{2}}} {cosxdx}−\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} {cosxdx}=\mathrm{2} \\ $$

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