Final Answer
Step-by-step Solution
Problem to solve:
Choose the solving method
We can solve the integral $\int\frac{1}{2\sin\left(x\right)\cos\left(x\right)}dx$ by applying the method Weierstrass substitution (also known as tangent half-angle substitution) which converts an integral of trigonometric functions into a rational function of $t$ by setting the substitution
Hence
Substituting in the original integral we get
Multiplying fractions $\frac{2t}{1+t^{2}} \times \frac{1-t^{2}}{1+t^{2}}$
Multiplying the fraction by $2$
Divide fractions $\frac{1}{\frac{4t\left(1-t^{2}\right)}{\left(1+t^{2}\right)^2}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$
A binomial squared (difference) is equal to the square of the first term, minus the double product of the first by the second, plus the square of the second term. In other words: $(a-b)^2=a^2-2ab+b^2$
- Square of the first term: $\left(1\right)^2 = 1^2$
- Double product of the first by the second: $2\left(1\right)\left(t^{2}\right) = 2\cdot 1t^{2}$
- Square of the second term: $\left(t^{2}\right)^2 = \left(t^{2}\right)^2$
Multiplying fractions $\frac{1+2t^{2}+t^{4}}{4t-4t^{3}} \times \frac{2}{1+t^{2}}$
Simplifying
Take out the constant $2$ from the integral
The trinomial $1+2t^{2}+t^{4}$ is a perfect square trinomial, because it's discriminant is equal to zero
Using the perfect square trinomial formula
Factoring the perfect square trinomial
Simplify the fraction $\frac{\left(t^{2}+1\right)^{2}}{\left(4t-4t^{3}\right)\left(1+t^{2}\right)}$ by $1+t^{2}$
Subtract the values $2$ and $-1$
Any expression to the power of $1$ is equal to that same expression
Any expression multiplied by $1$ is equal to itself
Factor the polynomial $4t-4t^{3}$ by it's GCF: $4t$
Rewrite the expression $\frac{1+2t^{2}+t^{4}}{\left(4t-4t^{3}\right)\left(1+t^{2}\right)}$ inside the integral in factored form
Take the constant $\frac{1}{4}$ out of the integral
Divide $1$ by $4$
Multiply $2$ times $\frac{1}{4}$
Take the constant $\frac{1}{4}$ out of the integral
Rewrite the fraction $\frac{1+t^{2}}{t\left(1-t^2\right)}$ in $2$ simpler fractions using partial fraction decomposition
Find the values for the unknown coefficients: $A, B, C$. The first step is to multiply both sides of the equation from the previous step by $t\left(1-t^2\right)$
Multiplying polynomials
Simplifying
Expand the polynomial
Assigning values to $t$ we obtain the following system of equations
Proceed to solve the system of linear equations
Rewrite as a coefficient matrix
Reducing the original matrix to a identity matrix using Gaussian Elimination
The integral of $\frac{1+t^{2}}{t\left(1-t^2\right)}$ in decomposed fraction equals
Expand the integral $\int\left(\frac{1}{t}+\frac{2t}{1-t^2}\right)dt$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately
Taking the constant ($2$) out of the integral
Simplifying
We can solve the integral $\int\frac{t}{1-t^2}dt$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $1-t^2$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part
Differentiate both sides of the equation $u=1-t^2$
Find the derivative
The derivative of a sum of two or more functions is the sum of the derivatives of each function
The derivative of the constant function ($1$) is equal to zero
The derivative of a function multiplied by a constant ($-1$) is equal to the constant times the derivative of the function
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
Now, in order to rewrite $dt$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above
Isolate $dt$ in the previous equation
Simplify the fraction $\frac{\frac{t}{u}}{-2t}$ by $t$
Take the constant $\frac{1}{-2}$ out of the integral
Substituting $u$ and $dt$ in the integral and simplify
The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$
The integral $\frac{1}{2}\int\frac{1}{t}dt$ results in: $\frac{1}{2}\ln\left(t\right)$
The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$
Replace $u$ with the value that we assigned to it in the beginning: $1-t^2$
The integral $-\frac{1}{2}\int\frac{1}{u}du$ results in: $-\frac{1}{2}\ln\left(1-t^2\right)$
Gather the results of all integrals
Replace $t$ with the value that we assigned to it in the beginning: $\tan\left(\frac{x}{2}\right)$
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$