Question

sin 4x sin 8x.dx
Integrate the function

Answer 1

HELLO DEAR,

GIVEN function is integral of (sin4x*sin8x).dx

we know:- sinAsinB = 1/2[cos(A-B)-cos(A+B)]

NOW, $\sf{\int{sin4x*sin8x}\,dx}$

=$\sf{\int{1/2[cos(-4x)-cos(12x)]}\,dx}$

=$\sf{\int{1/2[cos4x-cos12x]}\,dx}$

=$\sf{1/2[\frac{sin4x}{4}-\frac{sin12x}{12}]+c}$

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

Step-by-step explanation: