Question

sin x sin 2x sin 3x.dx
Integrate the function

Answer 1

HELLO DEAR,


Given function is integral of (sinxsin2xsin3x).dx


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


$\sf{1/2[\int{sinx(cos(-x)-cos3x)}]\,dx}$

$\sf{1/2[\int{sinxcosx-sinxcos5x}]\,dx}$

[tex]\sf{1/2[\int{\frac{sin2x}{2}}\,dx]-1/4\int{[sin6x+
sin(-4x)]}\,dx}[/tex]

[tex]\sf{\frac{-cos2x}{8}-\frac{1}{4}[\frac{-cos6x}{6}+ \frac{cos4x}{4}]+c}[/tex]

[tex]\sf{\frac{1}{8}[\frac{cos6x}{3}-\frac{cos4x}{2}-
cos2x]+c}[/tex]

I HOPE ITS HELP YOU DEAR,
THANKS