Integrate cos(4x)cos(6x)

\int\cos\left(4x\right)\cdot\cos\left(6x\right)dx

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Answer

$-\frac{1}{4}\sin\left(-2x\right)+\frac{1}{20}\sin\left(10x\right)+C_0$

Step by step solution

Problem

$\int\cos\left(4x\right)\cdot\cos\left(6x\right)dx$
1

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

$\int\frac{\cos\left(4x-6x\right)+\cos\left(6x+4x\right)}{2}dx$
2

Taking the constant out of the integral

$\frac{1}{2}\int\left(\cos\left(4x-6x\right)+\cos\left(6x+4x\right)\right)dx$
3

Adding $4x$ and $6x$

$\frac{1}{2}\int\left(\cos\left(\left(4-6\right)x\right)+\cos\left(10x\right)\right)dx$
4

Subtract the values $4$ and $-6$

$\frac{1}{2}\int\left(\cos\left(-2x\right)+\cos\left(10x\right)\right)dx$
5

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

$\frac{1}{2}\left(\int\cos\left(-2x\right)dx+\int\cos\left(10x\right)dx\right)$
6

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

$\frac{1}{2}\left(\int\cos\left(10x\right)dx-\frac{1}{2}\sin\left(-2x\right)\right)$
7

Apply the formula: $\int\cos\left(x\cdot a\right)dx$$=\frac{1}{a}\sin\left(x\cdot a\right)$, where $a=10$

$\frac{1}{2}\left(\frac{1}{10}\sin\left(10x\right)-\frac{1}{2}\sin\left(-2x\right)\right)$
8

Multiply $\left(\frac{1}{10}\sin\left(10x\right)+-\frac{1}{2}\sin\left(-2x\right)\right)$ by $\frac{1}{2}$

$\frac{1}{20}\sin\left(10x\right)-\frac{1}{4}\sin\left(-2x\right)$
9

Add the constant of integration

$-\frac{1}{4}\sin\left(-2x\right)+\frac{1}{20}\sin\left(10x\right)+C_0$

Answer

$-\frac{1}{4}\sin\left(-2x\right)+\frac{1}{20}\sin\left(10x\right)+C_0$

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Problem Analysis

Main topic:

Integration by substitution

Time to solve it:

0.26 seconds

Views:

256