"m" is a random number.
I`d appreciate a clear explanation.
2007-02-17 07:20:51 · 4 answers · asked by Uncle Rodri 1 in Science & Mathematics ➔ Mathematics
Sorry, i ment ∫(cos mx)^2 dx
2007-02-17 07:22:57 · update #1
∫ (cos(mx))^2 * dx
= ∫ (1 + cos(2mx) ) / 2 * dx
= 1/2 * [ ∫ (1 * dx) + ∫ (cos(2mx) * dx) ]
= 1/2 * [ x + { sin(2mx) / 2m } ] + C
= { 2mx + sin(2mx) } / 4m + C
2007-02-17 07:31:25 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Since cos 2x = 2 (cos x)^2 -1
so (cos mx)^2 = 1/2(1+cos 2mx)
so â« (cos mx )2 dx = â« 1/2 (1+ cos 2mx) . dx
= 1/2 [ 1 + 1/(2m) sin 2mx ] + C
( c is the integeration constant )
2007-02-17 15:31:12 · answer #2 · answered by a_ebnlhaitham 6 · 0⤊ 0⤋
Use the identity cos2y = 2(cosy)^2 - 1 in reverse to get an expression in cos2mx which is easier to integrate.
2007-02-17 15:27:06 · answer #3 · answered by mathsmanretired 7 · 1⤊ 0⤋
cos(mx)^2*x
ITT we use TI89
2007-02-17 15:27:10 · answer #4 · answered by Nate R 2 · 0⤊ 0⤋