Question

Integrate: cosx cos3x dx.

Answer 1

hope it helps you ...just think about these basic formulas in any suvch question

Answer 2

The value of the given integral is

$\int \cos \left(x\right)\cos \left(3x\right)dx=\frac{1}{2}\left(\frac{1}{4}\sin \left(4x\right)+\frac{1}{2}\sin \left(2x\right)\right)+C$

Step-by-step explanation:

The given integral is

$\int \cos \left(x\right)\cos \left(3x\right)dx$

Use the identity: $\cos \left(s\right)\cos \left(t\right)=\frac{\cos \left(s+t\right)+\cos \left(s-t\right)}{2}$

So, the integral becomes

$=\int \frac{\cos \left(x+3x\right)+\cos \left(x-3x\right)}{2}dx$

Take the constant out of the integral

$=\frac{1}{2}\cdot \int \cos \left(x+3x\right)+\cos \left(x-3x\right)dx$

Apply the sum rule of integral: $\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=\frac{1}{2}\left(\int \cos \left(x+3x\right)dx+\int \cos \left(x-3x\right)dx\right)$

Apply the formula for common integral of cosine function

$\frac{1}{2}\left(\frac{1}{4}\sin \left(4x\right)+\frac{1}{2}\sin \left(2x\right)\right)+C$

#Learn More:

Integrate the definite integral

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