Question

cos 2x cos 4x cos 6x.dx
Integrate the function

Answer 1

HELLO DEAR,


GIVEN function is integral of (cos2xcos4xcos6x).dx

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

now, intgral of (cos2xcos4xcos6x).dx

=[tex]\sf{\int{cos 2x[\frac{1}{2}(cos10x+cos2x)]}\,dx}[/tex]

=[tex]\sf{\int{\frac{1}{2}(cos2xcos10x+ cos^22x)}\,dx}[/tex]

=[tex]\sf{\int{\frac{1}{4}[cos12x+cos8x]}\,dx+
\int{\frac{1}{4}(1+cos4x)}\,dx}[/tex]

=[tex]\sf{\int{\frac{1}{4}[cos12x+cos8x+cos4x+
1}\,dx}[/tex]

= [tex]\sf{\frac{1}{4}[\frac{sin12}{12}+\frac{cos8x}{8}+ \frac{cos4x}{4}+x]+c}[/tex]


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

the

cos 2x cos 4x cos 6x.dx