Weierstrass Substitution Calculator

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Difficult Problems

Solved example of weierstrass substitution

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

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

$t=\tan\left(\frac{x}{2}\right)$

Hence

$\sin x=\frac{2t}{1+t^{2}},\:\cos x=\frac{1-t^{2}}{1+t^{2}},\:\mathrm{and}\:\:dx=\frac{2}{1+t^{2}}dt$

Substituting in the original integral we get

$\int\frac{1}{1-\left(\frac{1-t^{2}}{1+t^{2}}\right)+\frac{2t}{1+t^{2}}}\frac{2}{1+t^{2}}dt$

Combine $1-\left(\frac{1-t^{2}}{1+t^{2}}\right)+\frac{2t}{1+t^{2}}$ in a single fraction

$\int\frac{1}{\frac{2t+1+t^{2}}{1+t^{2}}+\frac{-\left(1-t^{2}\right)}{1+t^{2}}}\frac{2}{1+t^{2}}dt$

Multiplying fractions $\frac{1}{\frac{2t+1+t^{2}}{1+t^{2}}+\frac{-\left(1-t^{2}\right)}{1+t^{2}}} \times \frac{2}{1+t^{2}}$

$\int\frac{2}{\left(\frac{2t+1+t^{2}}{1+t^{2}}+\frac{-\left(1-t^{2}\right)}{1+t^{2}}\right)\left(1+t^{2}\right)}dt$

Add fraction's numerators with common denominators: $\frac{2t+1+t^{2}}{1+t^{2}}$ and $\frac{-\left(1-t^{2}\right)}{1+t^{2}}$

$\int\frac{2\left(1+t^{2}\right)}{\left(2t^{2}+2t\right)\left(1+t^{2}\right)}dt$

Simplify the fraction $\frac{2\left(1+t^{2}\right)}{\left(2t^{2}+2t\right)\left(1+t^{2}\right)}$ by $1+t^{2}$

$\int\frac{2}{2t^{2}+2t}dt$

Factor the denominator by $2$

$\int\frac{2}{2\left(t^{2}+t\right)}dt$

Cancel the fraction's common factor $2$

$\int\frac{1}{t^{2}+t}dt$

Simplifying

$\int\frac{1}{t^{2}+t}dt$

Factor the polynomial $t^{2}+t$ by it's GCF: $t$

$\int\frac{1}{t\left(t+1\right)}dt$

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

$\frac{1}{t\left(t+1\right)}=\frac{A}{t}+\frac{B}{t+1}$

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 $t\left(t+1\right)$

$1=t\left(t+1\right)\left(\frac{A}{t}+\frac{B}{t+1}\right)$

Multiplying polynomials

$1=\frac{tA\left(t+1\right)}{t}+\frac{tB\left(t+1\right)}{t+1}$

Simplifying

$1=A\left(t+1\right)+tB$

Expand the polynomial

$1=At+A+tB$

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

$\begin{matrix}1=A&\:\:\:\:\:\:\:(t=0) \\ 1=-B&\:\:\:\:\:\:\:(t=-1)\end{matrix}$

Proceed to solve the system of linear equations

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

Rewrite as a coefficient matrix

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

Reducing the original matrix to a identity matrix using Gaussian Elimination

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

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

$\int\left(\frac{1}{t}+\frac{-1}{t+1}\right)dt$

Expand the integral $\int\left(\frac{1}{t}+\frac{-1}{t+1}\right)dt$

$\int\frac{1}{t}dt+\int\frac{-1}{t+1}dt$

We can solve the integral $\int\frac{-1}{t+1}dt$ 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 $t+1$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=t+1$

Differentiate both sides of the equation $u=t+1$

$du=\frac{d}{dt}\left(t+1\right)$

Find the derivative

$\frac{d}{dt}\left(t+1\right)$

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{d}{dt}\left(t\right)+\frac{d}{dt}\left(1\right)$

The derivative of the constant function ($1$) is equal to zero

$\frac{d}{dt}\left(t\right)$

The derivative of the linear function is equal to $1$

$1$

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

$du=dt$

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

$\int\frac{1}{t}dt+\int\frac{-1}{u}du$

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$\ln\left|t\right|$

The integral $\int\frac{1}{t}dt$ results in: $\ln\left|t\right|$

$\ln\left|t\right|$

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$-\ln\left|u\right|$

Replace $u$ with the value that we assigned to it in the beginning: $t+1$

$-\ln\left|t+1\right|$

The integral $\int\frac{-1}{u}du$ results in: $-\ln\left|t+1\right|$

$-\ln\left|t+1\right|$

Gather the results of all integrals

$\ln\left|t\right|-\ln\left|t+1\right|$

Replace $t$ with the value that we assigned to it in the beginning: $\tan\left(\frac{x}{2}\right)$

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

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

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

Final Answer

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

Weierstrass Substitution Calculator

Get detailed solutions to your math problems with our Weierstrass Substitution 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!

Example

Solved Problems

Difficult Problems

Final Answer

Struggling with math?

Weierstrass Substitution Calculator

Get detailed solutions to your math problems with our Weierstrass Substitution 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!

Example

Solved Problems

Difficult Problems

Final Answer

Struggling with math?

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