Question

Find the integrals (primitives):
$\rm \displaystyle\int \cos 2x.\cos 4x.\cos 6x \ dx$

Answer 1

❤See in the given pic it is been solved:)

Answer 2

Answer:

$\frac{x}{4}+\frac{Sin4x}{16}+\frac{Sin8x}{32} +\frac{Sin12x}{48} +C$

Step-by-step explanation:

We have find the value of

$\int\ {Cos2x.Cos4x.Cos6x} \, dx$

=$\frac{1}{2} \int\ {(2Cos2x.Cos4x).Cos6x} \, dx$

=$\frac{1}{2} \int\ {(Cos6x+Cos2x).Cos6x} \, dx$

Since, we know the formula, 2CosA.CosB=Cos(A+B)+Cos(A-B)

=$\frac{1}{2} \int\ {Cos^{2}6x } \, dx +\frac{1}{2} \int\ {Cos2x.Cos6x} \, dx$

Again applying the formula, 2CosA.CosB=Cos(A+B)+Cos(A-B) and

Cos2x=2Cos²x-1

=$\frac{1}{4} \int\ {2Cos^{2}6x } \, dx +\frac{1}{4} \int\ {(Cos8x+Cos4x)} \, dx$

=$\frac{1}{4} \int\ {(1+Cox12x)} \, dx +\frac{1}{4} \int\ {(Cos8x+Cos4x)} \, dx$

=$\frac{1}{4} [x+\frac{Sin12x}{12}]+\frac{1}{4} [\frac{Sin8x}{8} +\frac{Sin4x}{4}]+C$

Where C is an integration constant.

=$\frac{x}{4}+\frac{Sin4x}{16}+\frac{Sin8x}{32} +\frac{Sin12x}{48} +C$