Solved example of weierstrass substitution
We can solve the integral $\int\frac{1}{1-\cos\left(x\right)+\sin\left(x\right)}dx$ 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
Add fraction's numerators with common denominators: $\frac{2t}{1+t^{2}}$ and $\frac{1-t^{2}}{1+t^{2}}$
Combine $1+\frac{2t-\left(1-t^{2}\right)}{1+t^{2}}$ in a single fraction
Divide fractions $\frac{1}{\frac{2t-\left(1-t^{2}\right)+1+t^{2}}{1+t^{2}}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$
Solve the product $-(1-t^{2})$
Multiplying fractions $\frac{1+t^{2}}{2t+2t^{2}} \times \frac{2}{1+t^{2}}$
Simplify the fraction $\frac{2\left(1+t^{2}\right)}{\left(2t+2t^{2}\right)\left(1+t^{2}\right)}$ by $1+t^{2}$
Factor the denominator by $2$
Cancel the fraction's common factor $2$
Simplifying
Factor the polynomial $t+t^{2}$ by it's GCF: $t$
Rewrite the fraction $\frac{1}{t\left(1+t\right)}$ in $2$ simpler fractions using partial fraction decomposition
Find the values for the unknown coefficients: $A, B$. The first step is to multiply both sides of the equation from the previous step by $t\left(1+t\right)$
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 $\frac{1}{t\left(1+t\right)}$ in decomposed fraction equals
Expand the integral $\int\left(\frac{1}{t}+\frac{-1}{1+t}\right)dt$
We can solve the integral $\int\frac{-1}{1+t}dt$ 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 $1+t$ 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 $u=1+t$
Find the derivative
The derivative of a sum of two functions is the sum of the derivatives of each function
The derivative of the constant function ($1$) is equal to zero
The derivative of the linear function is equal to $1$
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
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 $\int\frac{1}{t}dt$ results in: $\ln\left(t\right)$
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: $1+t$
The integral $\int\frac{-1}{u}du$ results in: $-\ln\left(1+t\right)$
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 $C$
Access detailed step by step solutions to thousands of problems, growing every day!