Soal integral

yang pinter-pinter, entah hanya ngerasa ato emang pinter beneran, tolong ya!

soal:
Tunjukkan bahwa:
a.

b.

tolong nih

Pake tulisan biasa aja ya, mas...:))
Aku rada gaptek kalo harus pake fitur penulisan yang sudah disediakan itu...

1. sin^4(x)

= [sin^2(x)]^2

= [(1 -cos(2x)/2)]^2

= 1/4[1 + cos^2(2x) - 2cos(2x)]

= 1/4[1 - 2cos(2x) + (1 + cos(4x)/2 ]

= (1/4) - (1/2)cos(2x) + (1/8) + (1/8)cos(4x)

= (3/8) - (1/2)cos(2x) + (1/8)cos(4x)

maka ∫sin^4(x)dx = 3/8∫dx - 1/2∫cos(2x) dx + 1/8∫cos(4x) dx

∫sin^4(x)dx = (3/8)x - (1/2)sin(2x)(1/2) + (1/8)sin(4x)(1/4) + C

∫sin^4(x)dx = (3/8)x - (1/4)sin(2x) + (1/32)sin(4x) + C



2. ∫ cos^4(x) dx = ∫ (cos^2x)^2 dx

karena cos^2x = (1/2)[1 + cos(2x)], maka menjadi

∫ {(1/2)[1 + cos(2x)]}^2 dx = ∫ (1/4)[1 + cos(2x)]^2 dx

∫ (1/4)[1 + 2cos(2x) + cos^2(2x)] dx

= (1/4) ∫ dx + (1/4) 2 ∫ cos(2x) + (1/4) ∫ cos^2(2x) dx

= (1/4) ∫ dx + (1/2) ∫ cos(2x) + (1/4) ∫ cos^2(2x) dx

= (1/4)x + (1/2) (1/2)sin(2x) + (1/4) ∫ cos^2(2x) dx

= (1/4)x + (1/4)sin(2x) + (1/4) ∫ cos^2(2x) dx

dari hasil di atas, maka

cos^2(2x) = (1/2){1 + cos[2(2x)]} = (1/2)[1 + cos(4x)]

oleh karena itu,

(1/4)x + (1/4)sin(2x) + (1/4) ∫ (1/2)[1 + cos(4x)] dx

= (1/4)x + (1/4)sin(2x) + (1/4)(1/2) ∫ [1 + cos(4x)] dx

= (1/4)x + (1/4)sin(2x) + (1/8) ∫ dx + (1/8) ∫ cos(4x) dx

= (1/4)x + (1/4)sin(2x) + (1/8)x + (1/8) (1/4)sin(4x) + C

= [(1 + 2)/8]x + (1/4)sin(2x) + (1/32)sin(4x) + C

Maka,

∫ cos^4(x) dx = (3/8)x + (1/4)sin(2x) + (1/32)sin(4x) + C


-dipi-

ih lihat jawabannya non dipi aja kau udah puyeng..... :D

@non dipi:
wohoho, TOP dah, tadi pagi untungnya aku sempet nanya temenku yg juara olim, dan hasilnya sama, repu deh bwt non!

Ampun God.......... puyeng liatnya... klasih reppu aja deh ke tante dipi dan den musthaf :D

Jadi kangen masa sma dl,

Krn guru Mat gw cinta bgt ma 'Integral' khususny 'Integral Parsial'..

Dohh, tiap hr pr Integral itu berlembar2 boo...
^^

pusing

wow mumet bener gan
<:||

waduh ga ngerti