About Snapxam Calculators Topics Contact us

# Integrate cos(4x)cos(6x)

Go!
1
2
3
4
5
6
7
8
9
0
x
y
(◻)
◻/◻
2

e
π
ln
log
lim
d/dx
d/dx
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

## 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

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

### Struggling with math?

Access detailed step by step solutions to millions of problems, growing every day!

### Main topic:

Integration by substitution

0.26 seconds

256