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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

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

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

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

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

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

Divide fractions $\frac{1}{\frac{2t-\left(1-t^{2}\right)+1+t^{2}}{1+t^{2}}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$

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

Solve the product $-(1-t^{2})$

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

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

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

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

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

Factor the denominator by $2$

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

Cancel the fraction's common factor $2$

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

Simplifying

$\int\frac{1}{t+t^{2}}dt$
6

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

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

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

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

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

Multiplying polynomials

$1=\frac{tA\left(1+t\right)}{t}+\frac{tB\left(1+t\right)}{1+t}$
10

Simplifying

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

Expand the polynomial

$1=A+At+tB$
12

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

$\begin{matrix}1=A&\:\:\:\:\:\:\:(t=0) \\ 1=-B&\:\:\:\:\:\:\:(t=-1)\end{matrix}$
13

Proceed to solve the system of linear equations

$\begin{matrix}1A & + & 0B & =1 \\ 0A & - & 1B & =1\end{matrix}$
14

Rewrite as a coefficient matrix

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

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 1 \\ 0 & 1 & -1\end{matrix}\right)$
16

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

$\int\left(\frac{1}{t}+\frac{-1}{1+t}\right)dt$
17

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

$\int\frac{1}{t}dt+\int\frac{-1}{1+t}dt$
18

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

$u=1+t$

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

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

Find the derivative

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

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

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

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 $1$

$1$
19

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

$du=dt$
20

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

$\ln\left(t\right)$
21

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: $1+t$

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

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

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

Gather the results of all integrals

$\ln\left(t\right)-\ln\left(1+t\right)$
24

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

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

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

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

Final Answer

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

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