Indefinite Integral?

1 ⤊

Indefinite Integral?

Integral: sin(6x)cos(11x)

2007-05-15 19:03:05 · 2 answers · asked by Mark S 1 in Science & Mathematics ➔ Mathematics

2 answers

Consider that sin(A+B) = sinAcosB + cosAsinB
sin(A-B) = sinAcosB - cosAsinB

So adding them
sin(A+B) + sin(A-B) = 2*sinAcosB

Now sin(6x)cos(11x) = (1/2) * [sin(17x) + sin(-5x)]
= (1/2) * [sin(17x) - sin(5x)]

That's a lot easier to integrate

2007-05-15 19:07:10 · answer #1 · answered by Dr D 7 · 1⤊ 0⤋

-5/170 * cos(17x) + 1/10 * cos(5x)

2007-05-16 02:06:57 · answer #2 · answered by gjmb1960 7 · 0⤊ 0⤋