Question

Integral tak tentu
1. ∫3 sin^2 dx

Answer 1

Verified answer

InTegraL Tak TentU

∫3 sin² x dx
= 3 ∫(1 - cos 2x)/2 dx
= 3(1/2 x - 1/4 ∫dsin 2x)
= 3/2 x - 3/4 sin 2x + C

Answer 2

Verified answer

IntEgral Tak TentU Trigonometri

cos 2x = 1 - 2 sin² x
2 sin² x= 1 - cos  2x
sin² x = 1/2 (1 -  cos 2x)

∫ 3 sin² x dx  = ∫ 3 ( 1/2)(1 -  cos 2x) dx
= 3/2  ∫ 1 - cos 2x  dx  
= (3/2)(x - 1/2 sin 2x) + c
= 3/2 x  - 3/4  sin 2x + c
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