# Factor by difference of squares Calculator

## Get detailed solutions to your math problems with our Factor by difference of squares step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here!

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### Difficult Problems

1

Solved example of factor by difference of squares

$\int\:\left(\frac{4\sin\:\left(x\right)}{\left(\left(4\cos\:\left(x\right)^2\right)-16\right)}\right)dx$
2

Taking the constant out of the integral

$4\int\frac{\sin\left(x\right)}{4\cos\left(x\right)^2-16}dx$
3

Solve the integral $\int\frac{\sin\left(x\right)}{4\cos\left(x\right)^2-16}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=\cos\left(x\right) \\ du=-\sin\left(x\right)dx\end{matrix}$
4

Isolate $dx$ in the previous equation

$\frac{du}{-\sin\left(x\right)}=dx$
5

Substituting $u$ and $dx$ in the integral and simplify

$4\int\frac{1}{-4u^2+16}du$
6

Factor the integral's denominator by $-4$

$4\int\frac{1}{-4\left(-4+u^2\right)}du$
7

Take the constant out of the integral

$4-\frac{1}{4}\int\frac{1}{-4+u^2}du$
8

Multiply $4$ times $-\frac{1}{4}$

$-\int\frac{1}{-4+u^2}du$
9

Rewrite the difference of squares $-4+u^2$ as the product of two conjugated binomials

$-\int\frac{1}{\left(2+u\right)\left(2-u\right)}du$
10

Rewrite the fraction $\frac{1}{\left(2+u\right)\left(2-u\right)}$ in $2$ simpler fractions using partial fraction decomposition

$\frac{1}{\left(2+u\right)\left(2-u\right)}=\frac{A}{2+u}+\frac{B}{2-u}$
11

Find the values of the unknown coefficients. The first step is to multiply both sides of the equation by $\left(2+u\right)\left(2-u\right)$

$1=\left(2+u\right)\left(2-u\right)\left(\frac{A}{2+u}+\frac{B}{2-u}\right)$
12

Multiplying polynomials

$1=\frac{A\left(2+u\right)\left(2-u\right)}{2+u}+\frac{B\left(2+u\right)\left(2-u\right)}{2-u}$
13

Simplifying

$1=A\left(2-u\right)+B\left(2+u\right)$
14

Expand the polynomial

$1=2A-Au+2B+Bu$
15

Assigning values to $u$ we obtain the following system of equations

$\begin{matrix}1=4A&\:\:\:\:\:\:\:(u=-2) \\ 1=4B&\:\:\:\:\:\:\:(u=2)\end{matrix}$
16

Proceed to solve the system of linear equations

$\begin{matrix}4A & + & 0B & =1 \\ 0A & + & 4B & =1\end{matrix}$
17

Rewrite as a coefficient matrix

$\left(\begin{matrix}4 & 0 & 1 \\ 0 & 4 & 1\end{matrix}\right)$
18

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & \frac{1}{4} \\ 0 & 1 & \frac{1}{4}\end{matrix}\right)$
19

The integral of $\frac{1}{\left(2+u\right)\left(2-u\right)}$ in decomposed fraction equals

$-\int\left(\frac{\frac{1}{4}}{2+u}+\frac{\frac{1}{4}}{2-u}\right)du$
20

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

$-\int\frac{\frac{1}{4}}{2+u}du-\int\frac{\frac{1}{4}}{2-u}du$
21

Apply the formula: $\int\frac{n}{ax+b}dx$$=\frac{n}{a}\ln\left|ax+b\right|, where a=-1, b=2, x=u and n=\frac{1}{4} -\int\frac{\frac{1}{4}}{2+u}du+\frac{1}{4}\ln\left|-u+2\right| 22 Substitute u back for it's value, \cos\left(x\right) -\int\frac{\frac{1}{4}}{2+u}du+\frac{1}{4}\ln\left|-\cos\left(x\right)+2\right| 23 Apply the formula: \int\frac{n}{x+b}dx$$=n\ln\left|x+b\right|$, where $b=2$, $x=u$ and $n=\frac{1}{4}$

$-\frac{1}{4}\ln\left|u+2\right|+\frac{1}{4}\ln\left|-\cos\left(x\right)+2\right|$
24

Substitute $u$ back for it's value, $\cos\left(x\right)$

$-\frac{1}{4}\ln\left|\cos\left(x\right)+2\right|+\frac{1}{4}\ln\left|-\cos\left(x\right)+2\right|$
25

As the integral that we are solving is an indefinite integral, when we finish we must add the constant of integration

$-\frac{1}{4}\ln\left|\cos\left(x\right)+2\right|+\frac{1}{4}\ln\left|-\cos\left(x\right)+2\right|+C_0$

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