∫4cos7xcos3x dx
2007-10-30 19:52:56 · 4 answers · asked by mikemallow 1 in Science & Mathematics ➔ Mathematics
2 cos 7x cos 3x = cos (7x-3x) + cos (7x+3x)
= cos 4x + cos 10x
so 4 cos 7x cos 3x = 2 cos 4x + 2 cos 10x
integrating we get (2 sin 4x)/4+ 2(sin 10x)/10
= (sin 4x)/2 + (sin 10x)/5
2007-10-30 20:02:59 · answer #1 · answered by Mein Hoon Na 7 · 1⤊ 1⤋
if you write for cos(7x) *cos(4x) = cos(ax) * cos(bx)
then is with the formula for products
cos(ax) * cos(bx) = 1/2[cos(a-b)x + cos(a+b)x)]
=1/2[cos(7-3)x + cos(7+3)x] = 1/2[cos(4x) + cos(10x)]
integral[cos(mx)dx = 1/m * sin(mx)
with m1=4 and m2=10 is
Integral[cos(4x)= 1/4 *sin(4x)
Integral[cos(10x)=1/10 *sin(10x)
write the factor 1/2 and 4 before the integrals
Integral[4*cos(7x) * cos(3x) dx = 4*[sin(4x)/8 + sin(10x)/20]
=sin(4x)/2 + sin(10x)/5
2007-10-30 22:13:32 · answer #2 · answered by Xenophon 3 · 0⤊ 0⤋
4 cos 7x cos 3x = (4)(1/2)[ cos 10x + cos 4x ]
I = 2 ∫ cos 10x + cos 4x dx
I = 2 [ sin 10x / 10 + sin 4x / 4 ] + C
I = (1/5) sin 10x + (1/2) sin 4x + C
2007-10-31 02:58:08 · answer #3 · answered by Como 7 · 0⤊ 0⤋
trig property
2cos(A)cos(B) = cos(A+B) + cos(A-B)
therefore,
∫4cos7xcos3x dx
= 2∫[cos(10x) + cos(4x)]dx
= (2/10)sin(10x) + (2/4)sin(4x)
= (1/5)sin(10x) + (1/2)sin(4x)
2007-10-30 20:22:21 · answer #4 · answered by bustedtaillights 4 · 0⤊ 0⤋