Integrasjon: Trigonometrisk substitusjon

Trigonometrisk substitusjon med sinus invers:

∫11−x2dx=sin⁡−1(x)+C\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:

∫1x2+1dx=tan⁡−1(x)+C\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?

∫11−x2dx\int \frac{1}{\sqrt{1-x^2}} dx

Substituerer u=sin−1⁡(x), dvs. x=sin⁡(u). Deriverer for å finne et uttrykk for dx:

x=sin⁡(u)⇒dxdu=cos⁡(u)∣⋅du⇒dx=cos⁡(u)  du\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:

∫11−x2dx=∫cos⁡(u)1−sin⁡2(u)du\int \frac{1}{\sqrt{1-x^2}}dx = \int \frac{\cos(u)}{\sqrt{1 - \sin^2(u) }} du

Husk at sin2⁡(u)+cos2⁡(u)=1 som gir cos2⁡(u)=.1–sin2⁡(u):

∫11−x2dx=∫cos⁡(u)cos⁡2(u)du=∫cos⁡(u)cos⁡(u)du=∫du=u+C\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):

∫11−x2dx=sin⁡−1(x)+C\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?

∫1x2+1dx\int \frac{1}{x^2 + 1} dx

Substituerer u=tan−1⁡(x), dvs. x=tan⁡(u). Deriverer tangens for å finne et uttrykk for dx:

x=tan⁡(u)⇒dxdu=tan⁡2(u)+1∣⋅du⇒dx=(tan⁡2(u)+1)  du\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:

∫1x2+1dx=∫tan⁡2(u)+1tan⁡2(u)+1du=∫du=u+C\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):

∫1x2+1dx=tan⁡−1(x)+C\int \frac{1}{x^2 + 1}dx = \tan^{-1}(x) + C

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

+ Eksempel: ∫19–x2dx

∫19−x2dx\int \frac{1}{\sqrt{9-x^2}}dx

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

9−x2=9(1−x29)=9⋅1−x232=31−(x3)2\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:

∫19−x2dx=13∫11−(x3)2dx\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:

∫11−x2dx=sin⁡−1x+C\int \frac{1}{\sqrt{1 - x^2}} dx = \sin^{-1} x + C

Substituerer u=x3. Deriverer for å finne et uttrykk for dx:

x3=u⇒13=dudx∣⋅3  dx⇒dx=3  du\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:

∫19−x2dx=13∫11−(x3)2dx⇒∫19−x2dx=13∫11−u2⋅3du⇒∫19−x2dx=∫11−u2du⇒∫19−x2dx=sin⁡−1u+C⇒∫19−x2dx=sin⁡−1(x3)+C\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=x3, dvs. sin−1⁡(x3)=u. Deriverer for å finne et uttrykk for dx:

x3=sin⁡u⇒13=cos⁡u  dudx∣⋅3  dx⇒dx=3cos⁡u  du\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:

∫19−x2dx=13∫11−(x3)2dx⇒∫19−x2dx=13∫11−sin⁡2u⋅3cos⁡u  du⇒∫19−x2dx=∫1cos⁡2u⋅cos⁡u  du⇒∫19−x2dx=∫1cos⁡u⋅cos⁡u  du⇒∫19−x2dx=∫du⇒∫19−x2dx=u+C⇒∫19−x2dx=sin⁡−1(x3)+C\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: ∫1x2+9dx

∫1x2+9dx\int \frac{1}{x^2 + 9}dx

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

x2+9=9(x29+1)=9(x232+1)=9((x3)2+1)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:

∫1x2+9dx=19∫1(x3)2+1dx\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:

∫1x2+1dx=tan⁡−1x+C\int \frac{1}{x^2 + 1} dx = \tan^{-1} x + C

Substituerer u=x3. Deriverer for å finne et uttrykk for dx:

x3=u⇒13=dudx∣⋅3  dx⇒dx=3  du\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:

∫1x2+9dx=19∫1(x3)2+1dx⇒∫1x2+1dx=19∫1u2+1⋅3du⇒∫1x2+1dx=13∫1u2+1du⇒∫1x2+1dx=13tan⁡−1u+C⇒∫1x2+1dx=13tan⁡−1(x3)+C\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=x3, dvs. tan−1⁡(x3)=u. Deriverer tangens for å finne et uttrykk for dx:

x3=tan⁡u⇒13dxdu=(1+tan⁡2u)∣⋅3  du⇒dx=3(1+tan⁡2u)  du\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:

∫1x2+9dx=19∫1(x3)2+1dx⇒∫1x2+9dx=19∫1tan⁡2u+1⋅3(1+tan⁡2u)du⇒∫1x2+9dx=13∫du⇒∫1x2+9dx=13u+C⇒∫1x2+9dx=13tan⁡−1(x3)+C\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: ∫1x2+2x+5dx

∫1x2+2x+5dx\int \frac{1}{x^2 + 2x + 5}dx

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

x2+2x+5=x2+2x+1+4=(x+1)2+4=4((x+1)24+1)=4((x+12)2+1)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:

∫1x2+9dx=14∫1(x+12)2+1dx\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:

∫1x2+1dx=tan⁡−1x+C\int \frac{1}{x^2 + 1} dx = \tan^{-1} x + C

Substituerer u=x+12. Deriverer for å finne et uttrykk for dx:

x+12=u⇒12=dudx∣⋅2  dx⇒dx=2  du\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:

∫1x2+2x+5dx=14∫1(x+12)2+1dx⇒∫1x2+2x+5dx=14∫1u2+1⋅2  du⇒∫1x2+2x+5dx=12∫1u2+1du⇒∫1x2+2x+5dx=12tan⁡−1u+C⇒∫1x2+2x+5dx=12tan⁡−1(x+12)+C\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=x+12, dvs. tan−1⁡(x+12)=u. Deriverer tangens for å finne et uttrykk for dx:

x+12=tan⁡u⇒12dxdu=(1+tan⁡2u)∣⋅2  du⇒dx=2(1+tan⁡2u)  du\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:

∫1x2+2x+5dx=14∫1(x+12)2+1dx⇒∫1x2+2x+5dx=14∫1tan⁡2u+1⋅2(1+tan⁡2u)du⇒∫1x2+2x+5dx=12∫du⇒∫1x2+2x+5dx=12u+C⇒∫1x2+2x+5dx=12tan⁡−1(x+12)+C\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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