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integral trigonometri

10
cos 2x = 1 - 2 sin² x
2 sin² x = 1 - cos 2x
(0..π) ∫ 2 sin² x dx = (0...π)∫ 1 - cos 2x  dx
= [x - 1/2 sin 2x ] (π...0)
= (π-0) - 1/2 (sin 2π - sin 0)
= π - 1/2 (0-0)
= π -0
= π

2.
(0..π/4) ∫ sin 2x - 2 cos 4x dx
= [ -1/2 cos 2x  - 1/2 sin 4x] (π/4 ...0)
= - 1/2 (cos π/2 - cos 0)  - 1/2 ( sin π - sin 0)
= - 1/2 (0 -1) - 1/2 (0-0)
=  1/2 - 0
=  1/2