Question

Tplong banget bantu plisssss

Answer 1

---> sin^4(x) 

=[ sin^2(x)]^2 

= ( (1-2cosx)/2 )^2 

= 1/4 (1 - 2cos2x +cos^2(x) ) 

= 1/4 (1 - 2cos2x + (1+cos4x)/2 ) 

= 1/4 - 1/2cos2x + 1/8 + 1/8cos4x 

= 3/8 - 1/2cos2x +1/8cos4x 

= 1/8 (3 - 4cos2x +cos4x) 

---> ∫sin^4(x)dx 
= 1/8∫(3 - 4cos2x + cos 4x) dx 

= 1/8 {3x - 4(sin2x)/2 +1/4sin4x} +C 

=> (3/8)x - (1/4)sin2x + (1/32)sin4x + C

(C = konstanta)

Semoga membantu^^