# Integral of cos(5x)cos(2x)

## \int\cos\left(5x\right)\cdot\cos\left(2x\right)dx

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$\frac{1}{6}\sin\left(3x\right)+\frac{1}{14}\sin\left(7x\right)+C_0$

## Step by step solution

Problem

$\int\cos\left(5x\right)\cdot\cos\left(2x\right)dx$
1

Applying the rule of the product of two cosines $\cos\left(a\right)\cdot\cos\left(b\right)=\frac{\cos\left(a+b\right)+\cos\left(a-b\right)}{2}$

$\int\frac{\cos\left(5x-2x\right)+\cos\left(2x+5x\right)}{2}dx$
2

Taking the constant out of the integral

$\frac{1}{2}\int\left(\cos\left(5x-2x\right)+\cos\left(2x+5x\right)\right)dx$
3

Adding $5x$ and $2x$

$\frac{1}{2}\int\left(\cos\left(\left(5-2\right)x\right)+\cos\left(7x\right)\right)dx$
4

Subtract the values $5$ and $-2$

$\frac{1}{2}\int\left(\cos\left(3x\right)+\cos\left(7x\right)\right)dx$
5

The integral of a sum of two or more functions is equal to the sum of their integrals

$\frac{1}{2}\left(\int\cos\left(3x\right)dx+\int\cos\left(7x\right)dx\right)$
6

Apply the formula: $\int\cos\left(x\cdot a\right)dx$$=\frac{1}{a}\sin\left(x\cdot a\right), where a=3 \frac{1}{2}\left(\frac{1}{3}\sin\left(3x\right)+\int\cos\left(7x\right)dx\right) 7 Apply the formula: \int\cos\left(x\cdot a\right)dx$$=\frac{1}{a}\sin\left(x\cdot a\right)$, where $a=7$

$\frac{1}{2}\left(\frac{1}{3}\sin\left(3x\right)+\frac{1}{7}\sin\left(7x\right)\right)$
8

Multiply $\left(\frac{1}{7}\sin\left(7x\right)+\frac{1}{3}\sin\left(3x\right)\right)$ by $\frac{1}{2}$

$\frac{1}{6}\sin\left(3x\right)+\frac{1}{14}\sin\left(7x\right)$
9

$\frac{1}{6}\sin\left(3x\right)+\frac{1}{14}\sin\left(7x\right)+C_0$

$\frac{1}{6}\sin\left(3x\right)+\frac{1}{14}\sin\left(7x\right)+C_0$

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### Main topic:

Integration by substitution

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