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Weierstrass Substitution Calculator

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1

Solved example of weierstrass substitution

$\int\frac{1}{1-cos\left(x\right)+sin\left(x\right)}dx$
2

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

$t=\tan\left(\frac{x}{2}\right)$
3

Hence

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

4

Substituting in the original integral we get

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

Multiplying fractions $\frac{1}{1-\frac{1-t^{2}}{1+t^{2}}+\frac{2t}{1+t^{2}}} \times \frac{2}{1+t^{2}}$

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

The least common multiple (LCM) of a sum of algebraic fractions consists of the product of the common factors with the greatest exponent, and the uncommon factors

$L.C.M.=1+t^{2}$

Obtained the least common multiple (LCM), we place it as the denominator of each fraction, and in the numerator of each fraction we add the factors that we need to complete

$\frac{1+t^{2}}{1+t^{2}}+\frac{-1+t^{2}}{1+t^{2}}+\frac{2t}{1+t^{2}}$

Combine and simplify all terms in the same fraction with common denominator $1+t^{2}$

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

Multiplying the fraction 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 $2$

$\int\frac{1}{t^{2}+t}dt$
5

Simplifying

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

Factor the polynomial $t^{2}+t$ by it's greatest common factor (GCF): $t$

$\frac{1}{t\left(t+1\right)}$
6

Rewrite the expression $\frac{1}{t^{2}+t}$ inside the integral in factored form

$\int\frac{1}{t\left(t+1\right)}dt$
7

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

$\frac{1}{t\left(t+1\right)}=\frac{A}{t}+\frac{B}{t+1}$
8

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

$1=t\left(t+1\right)\left(\frac{A}{t}+\frac{B}{t+1}\right)$
9

Multiplying polynomials

$1=\frac{t\left(t+1\right)A}{t}+\frac{t\left(t+1\right)B}{t+1}$
10

Simplifying

$1=\left(t+1\right)A+tB$
11

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

$\begin{matrix}1=A&\:\:\:\:\:\:\:(t=0) \\ 1=-B&\:\:\:\:\:\:\:(t=-1)\end{matrix}$
12

Proceed to solve the system of linear equations

$\begin{matrix}1A & + & 0B & =1 \\ 0A & - & 1B & =1\end{matrix}$
13

Rewrite as a coefficient matrix

$\left(\begin{matrix}1 & 0 & 1 \\ 0 & -1 & 1\end{matrix}\right)$
14

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 1 \\ 0 & 1 & -1\end{matrix}\right)$
15

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

$\int\left(\frac{1}{t}+\frac{-1}{t+1}\right)dt$
16

Expand the integral $\int\left(\frac{1}{t}+\frac{-1}{t+1}\right)dt$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately

$\int\frac{1}{t}dt+\int\frac{-1}{t+1}dt$
17

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 or more 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 $1$

$1$
18

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

$du=dt$
19

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, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$\ln\left(t\right)$
20

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: $t+1$

$-\ln\left(t+1\right)$
21

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

$-\ln\left(t+1\right)$
22

Gather the results of all integrals

$\ln\left(t\right)-\ln\left(t+1\right)$
23

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

$\ln\left(\tan\left(\frac{x}{2}\right)\right)-\ln\left(\tan\left(\frac{x}{2}\right)+1\right)$
24

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

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

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