Question

Integral dari sin^4 x dx adalah...

Answer 1

Panjang banget caranya

=>[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) 

jadi ∫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