Find the integration of sin2x dx from 0 to pi/2.

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Solution

$Let I = \int_{0}^{\frac{π}{2}} \sin 2 x dx = {[- \frac{\cos 2 x}{2}]}_{0}^{π / 2} = - \frac{1}{2} [\cos 2 \times \frac{π}{2} - \cos 2 \times 0] = - \frac{1}{2} [\cos π - \cos 0] = - \frac{1}{2} [- 1 - 1] = 1$

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Find the integration of sin2x dx from 0 to pi/2.

Let I = ∫0π2 sin 2x dx=-cos 2x20π/2=-12cos 2×π2 - cos 2×0=-12cos π - cos 0=-12-1 - 1=1

$Let I = \int_{0}^{\frac{π}{2}} \sin 2 x dx = {[- \frac{\cos 2 x}{2}]}_{0}^{π / 2} = - \frac{1}{2} [\cos 2 \times \frac{π}{2} - \cos 2 \times 0] = - \frac{1}{2} [\cos π - \cos 0] = - \frac{1}{2} [- 1 - 1] = 1$