Integrasjon: Trigonometrisk substitusjon

Trigonometrisk substitusjon med sinus invers:

\int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1}(x) + C

Sinus invers substitusjon brukes når nevneren er kvadratroten av noen andregradspolynom.

Trigonometrisk substitusjon med tangens invers:

\int \frac{1}{x^2 + 1} dx = \tan^{-1}(x) + C

Tangens invers substitusjon brukes når nevneren er noen andregradspolynom.

+ Hvordan virker substitusjon med sinus invers?

\int \frac{1}{\sqrt{1-x^2}} dx

Substituerer $\textcolor{red}{u = \sin^{-1}(x)}$, dvs. $x = \sin(u)$. Deriverer for å finne et uttrykk for $dx$:

\begin{aligned}
x & = \sin(u) \\
\Rightarrow \quad \frac{dx}{du} & = \cos(u) \qquad | \cdot du \\
\Rightarrow \quad dx & = \cos(u) \; du
\end{aligned}

Setter inn:

\int \frac{1}{\sqrt{1-x^2}}dx = \int \frac{\cos(u)}{\sqrt{1 - \sin^2(u) }} du

Husk at $\sin^2(u) + \cos^2(u) = 1$ som gir $\cos^2(u) =.1 – \sin^2(u)$:

\int \frac{1}{\sqrt{1-x^2}}dx = \int \frac{\cos(u)}{\sqrt{\cos^2(u)}} du = \int \frac{\cos(u)}{\cos(u)} du = \int du = u + C

Til sist bytter vi tilbake til $x$ ved å sette $u = \sin^{-1}(x)$:

\int \frac{1}{\sqrt{1-x^2}}dx = \sin^{-1}(x) + C

Og, vips, har vi vist formelen for substitusjon med sinus invers.

+ Hvordan virker substitusjon med tangens invers?

\int \frac{1}{x^2 + 1} dx

Substituerer $\textcolor{red}{u = \tan^{-1}(x)}$, dvs. $x = \tan(u)$. Deriverer tangens for å finne et uttrykk for $dx$:

\begin{aligned}
x & = \tan(u) \\
\Rightarrow \quad \frac{dx}{du} & = \tan^2(u)  + 1 \qquad | \cdot du \\
\Rightarrow \quad dx & = (\tan^2(u) + 1) \; du
\end{aligned}

Setter inn:

\int \frac{1}{x^2 + 1}dx = \int \frac{\tan^2(u) + 1}{\tan^2(u) + 1} du = \int du = u + C

Til sist bytter vi tilbake til $x$ ved å sette $u = \tan^{-1}(x)$:

\int \frac{1}{x^2 + 1}dx = \tan^{-1}(x) + C

Og, vips, har vi vist formelen for substitusjon med tangens invers.

+ Eksempel: $\int \frac{1}{\sqrt{9 – x^2}} dx$

\int \frac{1}{\sqrt{9-x^2}}dx

Først må vi gjøre noe for å få 1 – (uttrykk)2 under kvadratroten:

\sqrt{9 - x^2} = \sqrt{9\left(1 - \frac{x^2}{9}\right)} = \sqrt{9} \cdot \sqrt{1 - \frac{x^2}{3^2}}  = 3 \sqrt{1 - \left( \frac{x}{3}\right)^2 }

som gir:

\int \frac{1}{\sqrt{9 - x^2}} dx = \frac{1}{3} \int \frac{1}{\sqrt{1 - \left( \frac{x}{3} \right)^2 }} dx

+ Metode med formel

Vi ønsker å bruke formelen:

\int \frac{1}{\sqrt{1 - x^2}} dx = \sin^{-1} x + C

Substituerer $\textcolor{red}{u} = \textcolor{red}{\frac{x}{3}}$. Deriverer for å finne et uttrykk for $dx$:

\begin{aligned}
& \textcolor{red}{\frac{x}{3}} = \textcolor{red}{u} \\
\Rightarrow \quad & \frac{1}{3} = \frac{du}{dx} \qquad | \cdot 3 \; dx \\
\Rightarrow \quad & \textcolor{blue}{dx} = \textcolor{blue}{3\; du}
\end{aligned}

som gir:

\begin{aligned}
\int \frac{1}{\sqrt{9 - x^2}} dx & = \frac{1}{3} \int \frac{1}{\sqrt{1 - \left( \textcolor{red}{\frac{x}{3}} \right)^2 }} \textcolor{blue}{dx} \\
\Rightarrow \quad
\int \frac{1}{\sqrt{9 - x^2}} dx & = \frac{1}{3} \int \frac{1}{\sqrt{1 - \textcolor{red}{u}^2}} \cdot \textcolor{blue}{3 du} \\
\Rightarrow \quad
\int \frac{1}{\sqrt{9 - x^2}} dx & = \int \frac{1}{\sqrt{1-u^2}} du \\
\Rightarrow \quad
\int \frac{1}{\sqrt{9 - x^2}} dx &  = \sin^{-1}u + C \\
\Rightarrow \quad
\int \frac{1}{\sqrt{9 - x^2}} dx &  = \sin^{-1} \left( \frac{x}{3} \right) + C
\end{aligned}

+ Metode uten formel

Substituerer $\sin u = \frac{x}{3}$, dvs. $\textcolor{red}{\sin^{-1} \left(\frac{x}{3}\right)} = \textcolor{red}{u}$. Deriverer for å finne et uttrykk for $dx$:

\begin{aligned}
& \frac{x}{3} = \sin u \\
\Rightarrow \quad & \frac{1}{3} = \cos u \; \frac{du}{dx} \qquad | \cdot 3 \; dx \\
\Rightarrow \quad & \textcolor{blue}{dx} = \textcolor{blue}{3\cos u\; du}
\end{aligned}

som gir:

\begin{aligned}
\int \frac{1}{\sqrt{9 - x^2}} dx & = \frac{1}{3} \int \frac{1}{\sqrt{1 - \left( \textcolor{red}{\frac{x}{3}} \right)^2 }} \textcolor{blue}{dx} \\
\Rightarrow \quad
\int \frac{1}{\sqrt{9 - x^2}} dx & = \frac{1}{3} \int \frac{1}{\sqrt{1 - \textcolor{red}{\sin}^2 \textcolor{red}{u}}} \cdot \textcolor{blue}{3 \cos u \; du} \\ 
\Rightarrow \quad
\int \frac{1}{\sqrt{9 - x^2}} dx & = \int \frac{1}{\sqrt{\cos^2u}} \cdot \cos u \; du \\ 
\Rightarrow \quad
\int \frac{1}{\sqrt{9 - x^2}} dx & = \int \frac{1}{\cos u} \cdot \cos u \; du \\ 
\Rightarrow \quad
\int \frac{1}{\sqrt{9 - x^2}} dx & = \int du \\ 
\Rightarrow \quad
\int \frac{1}{\sqrt{9 - x^2}} dx & = u + C \\ 
\Rightarrow \quad
\int \frac{1}{\sqrt{9 - x^2}} dx & = \sin^{-1} \left( \frac{x}{3} \right) + C
\end{aligned}

Og, vips, har vi løst integralet.

+ Eksempel: $\int \frac{1}{x^2 + 9} dx$

\int \frac{1}{x^2 + 9}dx

Først må vi gjøre noe for å få (uttrykk)2 + 1 i nevneren:

x^2 + 9 =  9 \left( \frac{x^2}{9} + 1 \right) = 9 \left( \frac{x^2}{3^2} + 1 \right) = 9 \left( \left(\frac{x}{3}\right)^2 + 1 \right)

som gir:

\int \frac{1}{x^2 + 9} dx = \frac{1}{9} \int \frac{1}{\left( \frac{x}{3} \right)^2 + 1} dx

+ Metode med formel

Vi ønsker å bruke formelen:

\int \frac{1}{x^2 + 1} dx = \tan^{-1} x + C

Substituerer $\textcolor{red}{u} = \textcolor{red}{\frac{x}{3}}$. Deriverer for å finne et uttrykk for $dx$:

\begin{aligned}
& \textcolor{red}{\frac{x}{3}} = \textcolor{red}{u} \\
\Rightarrow \quad & \frac{1}{3} = \frac{du}{dx} \qquad | \cdot 3 \; dx \\
\Rightarrow \quad & \textcolor{blue}{dx} = \textcolor{blue}{3\; du}
\end{aligned}

som gir:

\begin{aligned}
\int \frac{1}{x^2 + 9} dx & = \frac{1}{9} \int \frac{1}{\left( \textcolor{red}{\frac{x}{3}} \right)^2  + 1} \textcolor{blue}{dx} \\
\Rightarrow \quad
\int \frac{1}{x^2 + 1} dx & = \frac{1}{9} \int \frac{1}{\textcolor{red}{u}^2 + 1} \cdot \textcolor{blue}{3 du} \\
\Rightarrow \quad
\int \frac{1}{x^2 + 1} dx & = \frac{1}{3} \int \frac{1}{u^2 + 1} du \\
\Rightarrow \quad
\int \frac{1}{x^2 + 1} dx &  = \frac{1}{3} \tan^{-1}u + C \\
\Rightarrow \quad
\int \frac{1}{x^2 + 1} dx &  = \frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right) + C
\end{aligned}

+ Metode uten formel

Substituerer $\tan u = \frac{x}{3}$, dvs. $\textcolor{red}{\tan^{-1} \left(\frac{x}{3}\right)} = \textcolor{red}{u}$. Deriverer tangens for å finne et uttrykk for $dx$:

\begin{aligned}
& \frac{x}{3} = \tan u \\
\Rightarrow \quad & \frac{1}{3} \frac{dx}{du}= ( 1+ \tan^2 u) \qquad | \cdot 3 \; du \\
\Rightarrow \quad & \textcolor{blue}{dx} = \textcolor{blue}{3(1 + \tan^2 u)\; du}
\end{aligned}

som gir:

\begin{aligned}
\int \frac{1}{x^2 + 9} dx & = \frac{1}{9} \int \frac{1}{\left( \textcolor{red}{\frac{x}{3}} \right)^2  + 1} \textcolor{blue}{dx} \\
\Rightarrow \quad
\int \frac{1}{x^2 + 9} dx & = \frac{1}{9} \int \frac{1}{\textcolor{red}{\tan}^2 \textcolor{red}{u} + 1} \cdot \textcolor{blue}{3 (1 + \tan^2u) du} \\ 
\Rightarrow \quad
\int \frac{1}{x^2 + 9} dx & = \frac{1}{3} \int du \\ 
\Rightarrow \quad
\int \frac{1}{x^2 + 9} dx & = \frac{1}{3} u + C \\ 
\Rightarrow \quad
\int \frac{1}{x^2 + 9} dx & = \frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right) + C
\end{aligned}

Og, vips, har vi løst integralet.

+ Eksempel: $\int \frac{1}{x^2 + 2x + 5} dx$

\int \frac{1}{x^2 + 2x + 5}dx

Først må vi gjøre noe for å få (uttrykk)2 + 1 i nevneren:

x^2 + 2x + 5 =  x^2 + 2x + 1 + 4 = (x+1)^2 + 4 = 4 \left( \frac{(x+1)^2}{4} + 1 \right) = 4 \left( \left(\frac{x+1}{2}\right)^2 + 1 \right)

som gir:

\int \frac{1}{x^2 + 9} dx = \frac{1}{4} \int \frac{1}{\left( \frac{x+1}{2} \right)^2 + 1} dx

+ Metode med formel

Vi ønsker å bruke formelen:

\int \frac{1}{x^2 + 1} dx = \tan^{-1} x + C

Substituerer $\textcolor{red}{u} = \textcolor{red}{\frac{x+1}{2}}$. Deriverer for å finne et uttrykk for $dx$:

\begin{aligned}
& \textcolor{red}{\frac{x+1}{2}} = \textcolor{red}{u} \\
\Rightarrow \quad & \frac{1}{2} = \frac{du}{dx} \qquad | \cdot 2 \; dx \\
\Rightarrow \quad & \textcolor{blue}{dx} = \textcolor{blue}{2\; du}
\end{aligned}

som gir:

\begin{aligned}
\int \frac{1}{x^2 + 2x + 5} dx & = \frac{1}{4} \int \frac{1}{\left( \textcolor{red}{\frac{x+1}{2}} \right)^2  + 1} \textcolor{blue}{dx} \\
\Rightarrow \quad
\int \frac{1}{x^2 + 2x + 5} dx & = \frac{1}{4} \int \frac{1}{\textcolor{red}{u}^2 + 1} \cdot \textcolor{blue}{2 \; du} \\
\Rightarrow \quad
\int \frac{1}{x^2 + 2x + 5} dx & = \frac{1}{2} \int \frac{1}{u^2 + 1} du \\
\Rightarrow \quad
\int \frac{1}{x^2 + 2x + 5} dx &  = \frac{1}{2} \tan^{-1}u + C \\
\Rightarrow \quad
\int \frac{1}{x^2 + 2x + 5} dx &  = \frac{1}{2} \tan^{-1} \left( \frac{x+1}{2} \right) + C
\end{aligned}

+ Metode uten formel

Substituerer $\tan u = \frac{x+1}{2}$, dvs. $\textcolor{red}{\tan^{-1} \left(\frac{x+1}{2}\right)} = \textcolor{red}{u}$. Deriverer tangens for å finne et uttrykk for $dx$:

\begin{aligned}
& \frac{x+1}{2} = \tan u \\
\Rightarrow \quad & \frac{1}{2} \frac{dx}{du} = ( 1+ \tan^2 u) \qquad | \cdot 2 \; du \\
\Rightarrow \quad & \textcolor{blue}{dx} = \textcolor{blue}{2(1 + \tan^2 u)\; du}
\end{aligned}

som gir:

\begin{aligned}
\int \frac{1}{x^2 + 2x + 5} dx & = \frac{1}{4} \int \frac{1}{\left( \textcolor{red}{\frac{x+1}{2}} \right)^2  + 1} \textcolor{blue}{dx} \\
\Rightarrow \quad
\int \frac{1}{x^2 + 2x + 5} dx & = \frac{1}{4} \int \frac{1}{\textcolor{red}{\tan}^2 \textcolor{red}{u} + 1} \cdot \textcolor{blue}{2 (1 + \tan^2u) du} \\ 
\Rightarrow \quad
\int \frac{1}{x^2 + 2x + 5} dx & = \frac{1}{2} \int du \\ 
\Rightarrow \quad
\int \frac{1}{x^2 + 2x + 5} dx & = \frac{1}{2} u + C \\ 
\Rightarrow \quad
\int \frac{1}{x^2 + 2x + 5} dx & = \frac{1}{2} \tan^{-1} \left( \frac{x+1}{2} \right) + C
\end{aligned}

Og, vips, har vi løst integralet.

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