Integration through substitution

The integration by substitution or substitution rule is an important method in the integral calculus to primitive functions and definite integrals to calculate. By introducing a new integration variable, part of the integrand is replaced in order to simplify the integral and thus ultimately lead it back to a known or more easily manageable integral.

The chain rule from differential calculus is the basis of the substitution rule. Its equivalent for integrals over multidimensional functions is the transformation theorem , which, however, presupposes a bijective substitution function.

Statement of the substitution rule

Let be a real interval , a continuous function and continuously differentiable . Then ${\displaystyle I}$ ${\displaystyle f\colon I\to \mathbb {R} }$ ${\displaystyle \varphi \colon [a,b]\to I}$

${\displaystyle \int _{a}^{b}f(\varphi (t))\cdot \varphi '(t)\,\mathrm {d} t=\int _{\varphi (a)}^{\varphi (b)}f(x)\,\mathrm {d} x.}$

proof

Let be an antiderivative of . According to the chain rule , the derivative of the composite function applies ${\displaystyle F}$ ${\displaystyle f}$ ${\displaystyle F\circ \varphi }$

${\displaystyle (F\circ \varphi )'(t)=F'(\varphi (t))\cdot \varphi '(t)=f(\varphi (t))\cdot \varphi '(t).}$

By applying the main theorem of differential and integral calculus twice , one obtains the substitution rule:

${\displaystyle {\begin{aligned}\int _{a}^{b}f(\varphi (t))\cdot \varphi '(t)\,\mathrm {d} t&{}=(F\circ \varphi )(b)-(F\circ \varphi )(a)\\&{}=F(\varphi (b))-F(\varphi (a))\\&{}=\int _{\varphi (a)}^{\varphi (b)}f(x)\,\mathrm {d} x\end{aligned}}}$

application

We consider:

${\displaystyle \int _{a}^{b}f(\varphi (t))\cdot \varphi '(t)\,\mathrm {d} t}$

The goal is to simplify the partial term of the integrand to the integration variable. This is done by applying the substitution rule. To do this, you first multiply the integrand by and then, in a second step, replace the integration variable with everywhere . In a final step, the integration limits and are replaced by or . ${\textstyle \varphi (t)}$ ${\textstyle t}$ ${\textstyle {1 \over \varphi '(t)}}$ ${\textstyle t}$ ${\textstyle \varphi ^{-1}(t)}$ ${\textstyle a}$ ${\textstyle b}$ ${\textstyle \varphi (a)}$ ${\textstyle \varphi (b)}$

So you educate

${\displaystyle {\begin{aligned}\int _{\varphi (a)}^{\varphi (b)}f(\varphi (\varphi ^{-1}(t))){\frac {\varphi '(\varphi ^{-1}(t))}{\varphi '(\varphi ^{-1}(t))}}\,\mathrm {d} t=\int _{\varphi (a)}^{\varphi (b)}f(t)\,\mathrm {d} t\end{aligned}}}$

For the sake of clarity, in practice one often goes to a new integration variable via z. B. from to . Then the inverse function reads and the differential becomes from to and one obtains the formally equivalent expression: ${\textstyle t}$ ${\textstyle x}$ ${\textstyle \varphi ^{-1}(x)}$ ${\textstyle \mathrm {d} t}$ ${\textstyle \mathrm {d} x}$

${\displaystyle {\begin{aligned}=\int _{\varphi (a)}^{\varphi (b)}f(x)\,\mathrm {d} x\end{aligned}}}$

Once you have found the antiderivative , you can evaluate it directly with the limits and or form the antiderivative to the original integrand as . ${\textstyle F}$ ${\textstyle \varphi (a)}$ ${\textstyle \varphi (b)}$ ${\textstyle F\circ \varphi }$

We can do the same backwards and apply the substitution rule

${\displaystyle {\begin{aligned}\int _{\varphi (a)}^{\varphi (b)}f(t)\,\mathrm {d} t\end{aligned}}}$

on. Then the integration variable has to be replaced by the term of and then multiplies the integrand by . Finally, one applies to the integration limits. ${\textstyle t}$ ${\textstyle \varphi (t)}$ ${\textstyle \varphi '(t)}$ ${\textstyle \varphi ^{-1}}$

Substitution of a definite integral

example 1

Calculating the integral

${\displaystyle \int _{0}^{a}\sin(2x)\,\mathrm {d} x}$

for any real number : Through the substitution we get , therefore , and thus: ${\displaystyle a>0}$ ${\displaystyle t=\varphi (x)=2x}$ ${\displaystyle \mathrm {d} t=\varphi '(x)\,\mathrm {d} x=2\,\mathrm {d} x}$ ${\displaystyle \mathrm {d} x={\tfrac {1}{2}}{\mathrm {d} t}}$

${\displaystyle \int _{0}^{a}\sin(2x)\,\mathrm {d} x=\int _{\varphi (0)}^{\varphi (a)}\sin(t)\,{\frac {1}{2}}{\mathrm {d} t}=\int _{0}^{2a}\sin(t)\,{\frac {1}{2}}{\mathrm {d} t}={\frac {1}{2}}\int _{0}^{2a}\sin(t)\,\mathrm {d} t}$

${\displaystyle ={\frac {1}{2}}[-\cos(t)]_{0}^{2a}={\frac {1}{2}}(-\cos(2a)+\cos(0))={\frac {1}{2}}(1-\cos(2a))}$.

Example 2

Calculating the integral

${\displaystyle \int _{0}^{2}x\cos \left(x^{2}+1\right)\,\mathrm {d} x}$:

The substitution gives , therefore , and thus ${\displaystyle t=\varphi (x)=x^{2}+1}$ ${\displaystyle \mathrm {d} t=2x\,\mathrm {d} x}$ ${\displaystyle x\,\mathrm {d} x={\tfrac {1}{2}}\mathrm {d} t}$

${\displaystyle \int _{0}^{2}x\cos \left(x^{2}+1\right)\,\mathrm {d} x={\frac {1}{2}}\int _{1}^{5}\cos(t)\,\mathrm {d} t={\frac {1}{2}}\left(\sin(5)-\sin(1)\right)}$.

So it is replaced by and by . The lower limit of the integral is converted into and the upper limit into . ${\displaystyle x^{2}+1}$ ${\displaystyle t}$ ${\displaystyle x\,\mathrm {d} x}$ ${\displaystyle {\tfrac {1}{2}}\,\mathrm {d} t}$ ${\displaystyle x=0}$ ${\displaystyle \varphi (0)=0^{2}+1=1}$ ${\displaystyle x=2}$ ${\displaystyle \varphi (2)=2^{2}+1=5}$

Example 3

For calculating the integral

${\displaystyle \int _{0}^{1}{\sqrt {1-x^{2}}}\,\mathrm {d} x}$

you can , so substitute. From this it follows . With you get ${\displaystyle x=\sin(t)}$ ${\displaystyle t=\arcsin(x)}$ ${\displaystyle \mathrm {d} x=\cos(t)\,\mathrm {d} t}$ ${\displaystyle {\sqrt {1-\sin ^{2}(t)}}=|{\cos(t)}|}$

${\displaystyle \int _{0}^{1}{\sqrt {1-x^{2}}}\,\mathrm {d} x=\int _{0}^{\frac {\pi }{2}}{\sqrt {1-\sin ^{2}(t)}}\cos(t)\,\mathrm {d} t=\int _{0}^{\frac {\pi }{2}}\cos ^{2}(t)\,\mathrm {d} t}$.

The result can be with partial integration or with the trigonometric formula

${\displaystyle \cos ^{2}(t)={\frac {1+\cos(2t)}{2}}}$

and a further substitution. It turns out

${\displaystyle \int _{0}^{1}{\sqrt {1-x^{2}}}\,\mathrm {d} x=\left[{\frac {t}{2}}+{\frac {1}{4}}\sin(2t)\right]_{0}^{\frac {\pi }{2}}={\frac {\pi }{4}}}$.

Substitution of an indefinite integral

Requirements and procedure

Under the above conditions applies

${\displaystyle \int f(x)\,\mathrm {d} x=\int f(\varphi (t))\cdot \varphi '(t)\,\mathrm {d} t.}$

After determining an antiderivative of the substituted function, the substitution is reversed and an antiderivative of the original function is obtained.

example 1

By completing the square and subsequent substitution , one obtains ${\displaystyle t=x+1}$ ${\displaystyle \mathrm {d} x=\mathrm {d} t}$

${\displaystyle \int {\frac {1}{x^{2}+2x+2}}\,\mathrm {d} x=\int {\frac {1}{(x+1)^{2}+1}}\,\mathrm {d} x=\int {\frac {1}{t^{2}+1}}\,\mathrm {d} t=\arctan(t)+C=\arctan(x+1)+C}$

Example 2

With the substitution one obtains ${\displaystyle t=x^{2},\mathrm {d} t=2x\,\mathrm {d} x}$

${\displaystyle \int x\,\cos \left(x^{2}\right)\,\mathrm {d} x={\frac {1}{2}}\int 2x\cos \left(x^{2}\right)\,\mathrm {d} x={\frac {1}{2}}\int \cos(t)\,\mathrm {d} t={\frac {1}{2}}\left(\sin(t)+C'\right)={\frac {1}{2}}\sin \left(x^{2}\right)+C}$

Note that the substitution is only for or only for strictly monotonic. ${\displaystyle x\geq 0}$ ${\displaystyle x\leq 0}$

Special cases of substitution

Linear substitution

Integrals with linear chaining can be calculated as follows: If is an antiderivative of , then holds ${\displaystyle F}$ ${\displaystyle f}$

${\displaystyle \int f(ax+b)\,\mathrm {d} x={\frac {1}{a}}F(ax+b)+C}$if . ${\displaystyle a\neq 0}$

For example

${\displaystyle \int e^{3x+1}\,\mathrm {d} x={\frac {1}{3}}e^{(3x+1)}+C}$,

there and . ${\displaystyle f(x)=e^{x}=F(x)}$ ${\displaystyle a=3}$

Logarithmic integration

Integrals where the integrand is a fraction, the numerator of which is the derivative of the denominator, can be solved very easily with the help of logarithmic integration:

${\displaystyle \int {\frac {f'(x)}{f(x)}}\,\mathrm {d} x=\ln |f(x)|+C\quad \left(f(x)\neq 0\right)}$.

This corresponds to a special case of the substitution method with . ${\displaystyle t=f(x)}$

For example

${\displaystyle \int {\frac {x}{x^{2}+1}}\,\mathrm {d} x={\frac {1}{2}}\int {\frac {2x}{x^{2}+1}}\,\mathrm {d} x={\frac {1}{2}}\ln(x^{2}+1)+C}$,

since the derivative has. ${\displaystyle f(x)=x^{2}+1}$ ${\displaystyle f'(x)=2x}$

Euler's substitution

According to Bernoulli's theorem, all integrals of the type

${\displaystyle \int {\sqrt {ax^{2}+bx+c}}\;\mathrm {d} x}$

and

${\displaystyle \int {\frac {\mathrm {d} x}{\sqrt {ax^{2}+bx+c}}}}$

integrate elementary.

Example:

${\displaystyle \int {\frac {\mathrm {d} x}{\sqrt {x^{2}+1}}}}$

By substituting therefore , , and results ${\displaystyle t-x={\sqrt {x^{2}+1}}}$ ${\displaystyle t^{2}-2tx=1}$ ${\displaystyle x={\tfrac {t}{2}}-{\tfrac {1}{2t}}}$ ${\displaystyle t-x={\tfrac {t}{2}}+{\tfrac {1}{2t}}}$ ${\displaystyle \mathrm {d} x=\left({\tfrac {1}{2}}+{\tfrac {1}{2t^{2}}}\right)\mathrm {d} t}$

${\displaystyle \int {\frac {\mathrm {d} x}{\sqrt {x^{2}+1}}}=\int {\frac {{\frac {1}{2}}+{\frac {1}{2t^{2}}}}{{\frac {t}{2}}+{\frac {1}{2t}}}}\mathrm {d} t=\int {\frac {\mathrm {d} t}{t}}=\ln t+C=\ln \left(x+{\sqrt {x^{2}+1}}\right)+C}$.

See also

• Partial integration for another important rule for calculating integrals,
• Weierstraß substitution for certain functions that contain trigonometric functions .

literature

• Harro Heuser: Textbook of Analysis. Part 1 , 5th edition, BG Teubner, Stuttgart 1988, ISBN 3-519-42221-2 , p. 464
• Konrad Königsberger: Analysis 1 , Springer, Berlin 1992, ISBN 3-540-55116-6 , pp. 200-201

Web links

• Simple explanation / examples for the substitution rule
• State education server BW: Method of linear substitution with detailed example and exercises / solutions
• Video: substitution rule . Jörn Loviscach 2011, made available by the Technical Information Library (TIB), doi : 10.5446 / 9911 .
• Video: Integration through substitution, finger exercise . Jörn Loviscach 2013, made available by the Technical Information Library (TIB), doi : 10.5446 / 10142 .
• Video: three ways to integrate through substitution . Jörn Loviscach 2013, made available by the Technical Information Library (TIB), doi : 10.5446 / 10144 .
• Video: partial integration, substitution rule, integration by partial fraction decomposition . Jörn Loviscach 2012, made available by the Technical Information Library (TIB), doi : 10.5446 / 9987 .
• Video: Examples of partial integration, substitution rule, integration through partial fraction decomposition . Jörn Loviscach 2012, made available by the Technical Information Library (TIB), doi : 10.5446 / 9988 .