Integrarea prin schimbul de variabilă este o metodă folosită pentru a simplifica o integrală prin introducerea unei noi variabile, ceea ce face calculul mai ușor. Aceasta implică următorii pași:
Exemplu
Să se calculeze integrală:
\[ \int \sin(4x + 3) \, dx \]
Aplicăm schimbul de variabilă:
- Notăm: \(t = 4x + 3\).
- Derivăm: \( dt = 4dx \implies dx = \frac{1}{4}dt\).
- Substituim în integrală:
\[ \int \sin(4x + 3) \, dx = \int \sin t \cdot \frac{1}{4} \, dt = \frac{1}{4} \int \sin t \, dt \]
Rezolvăm integrală în funcție de \(t\):
\[ \frac{1}{4} \int \sin t \, dt = \frac{1}{4} (-\cos t) + C = -\frac{1}{4} \cos t + C \]
Revenim la variabila inițială \(x\):
\[ -\frac{1}{4} \cos t + C = -\frac{1}{4} \cos(4x + 3) + C \]
1
Calculați: \(\displaystyle\int \sqrt{2x+7} \, dx\)
Răspuns: \(\displaystyle \frac{1}{3}(2x + 7)^{3/2} + C \)
2
Calculați: \(\displaystyle\int e^{5x-2} \, dx\)
Răspuns: \(\displaystyle \frac{1}{5} e^{5x - 2} + C \)
3
Calculați: \(\displaystyle\int \cos(9x+4) \, dx\)
Răspuns: \(\displaystyle \frac{1}{9} \sin(9x + 4) + C \)
4
Calculați: \(\displaystyle\int (-4x+3)^{19} \, dx\)
Răspuns: \(\displaystyle \frac{-1}{80}(-4x + 3)^{20} + C \)
5
Calculați: \(\displaystyle\int \frac{1}{(-7x+5)^2} \, dx\)
Răspuns: \(\displaystyle \frac{1}{7(-7x+5)} + C \)
6
Calculați: \(\displaystyle\int \frac{1}{\sqrt{6x-8}} \, dx\)
Răspuns: \(\displaystyle \frac{1}{3} \sqrt{6x - 8} + C \)
7
Calculați: \(\displaystyle\int 2^{5x-3} \, dx\)
Răspuns: \(\displaystyle \frac{1}{5\ln 2} \cdot 2^{5x - 3} + C \)
8
Calculați: \(\displaystyle\int \frac{1}{\cos^2(8x+1)} \, dx\)
Răspuns: \(\displaystyle \frac{1}{8} \operatorname{tg}(8x + 1) + C \)
9
Calculați: \(\displaystyle\int \frac{x^5}{\sqrt{x^6+4}} \, dx\)
Răspuns: \(\displaystyle \frac{1}{3} \sqrt{x^6 + 4} + C \)
10
Calculați: \(\displaystyle\int \frac{2x+3}{(x^2+3x-1)^2} \, dx\)
Răspuns: \(\displaystyle \frac{-1}{x^2 + 3x - 1} + C \)
11
Calculați: \(\displaystyle\int \frac{x}{x^2+4} \, dx\)
Răspuns: \(\displaystyle \frac{1}{2} \ln(x^2 + 4) + C \)
12
Calculați: \(\displaystyle\int \frac{1}{2\sqrt{x}(\sqrt{x}+1)} \, dx\)
Răspuns: \(\displaystyle \ln(\sqrt{x} + 1) + C \)
13
Calculați: \(\displaystyle\int \frac{\cos x}{\sin^2 x + 3} \, dx\)
Răspuns: \(\displaystyle \frac{1}{\sqrt{3}} \operatorname{arctg}\left( \frac{\sin x}{\sqrt{3}} \right) + C \)
14
Calculați: \(\displaystyle\int \frac{x}{\sqrt{3+x^2}} \, dx\)
Răspuns: \( \sqrt{3 + x^2} + C \)
15
Calculați: \(\displaystyle\int e^{x^3} \cdot x^2 \, dx\)
Răspuns: \(\displaystyle \frac{1}{3} e^{x^3} + C \)
16
Calculați: \(\displaystyle\int \frac{7^{\sqrt{x}}}{\sqrt{x}} \, dx\)
Răspuns: \(\displaystyle \frac{2 \cdot 7^{\sqrt{x}}}{\ln 7} + C \)
17
Calculați: \(\displaystyle\int \frac{4}{3}\sqrt[3]{5x+7} \, dx\)
Răspuns: \(\displaystyle \frac{1}{5} (5x + 7)^{4/3} + C \)
18
Calculați: \(\displaystyle\int \frac{1}{9x^2+16} \, dx\)
Răspuns: \(\displaystyle \frac{1}{12} \operatorname{arctg}\left( \frac{3x}{4} \right) + C \)
19
Calculați: \(\displaystyle\int \frac{e^x}{e^{2x}+16} \, dx\)
Răspuns: \(\displaystyle \frac{1}{4} \operatorname{arctg}\left( \frac{e^x}{4} \right) + C \)
20
Calculați: \(\displaystyle\int \frac{\cos x}{\sin^2 x - 3} \, dx\)
Răspuns: \(\displaystyle \frac{1}{2\sqrt{3}} \ln\left| \frac{\sin x - \sqrt{3}}{\sin x + \sqrt{3}} \right| + C \)
1
\(\displaystyle \frac{1}{3}(2x + 7)^{3/2} + C \)
2
\(\displaystyle \frac{1}{5} e^{5x - 2} + C \)
3
\(\displaystyle \frac{1}{9} \sin(9x + 4) + C \)
4
\(\displaystyle \frac{-1}{80}(-4x + 3)^{20} + C \)
5
\(\displaystyle \frac{1}{7(-7x+5)} + C \)
6
\(\displaystyle \frac{1}{3} \sqrt{6x - 8} + C \)
7
\(\displaystyle \frac{1}{5\ln 2} \cdot 2^{5x - 3} + C \)
8
\(\displaystyle \frac{1}{8} \operatorname{tg}(8x + 1) + C \)
9
\(\displaystyle \frac{1}{3} \sqrt{x^6 + 4} + C \)
10
\(\displaystyle \frac{-1}{x^2 + 3x - 1} + C \)
11
\(\displaystyle \frac{1}{2} \ln(x^2 + 4) + C \)
12
\(\displaystyle \ln(\sqrt{x} + 1) + C \)
13
\(\displaystyle \frac{1}{\sqrt{3}} \operatorname{arctg}\left( \frac{\sin x}{\sqrt{3}} \right) + C \)
14
\( \sqrt{3 + x^2} + C \)
15
\(\displaystyle \frac{1}{3} e^{x^3} + C \)
16
\(\displaystyle \frac{2 \cdot 7^{\sqrt{x}}}{\ln 7} + C \)
17
\(\displaystyle \frac{1}{5} (5x + 7)^{4/3} + C \)
18
\(\displaystyle \frac{1}{12} \operatorname{arctg}\left( \frac{3x}{4} \right) + C \)
19
\(\displaystyle \frac{1}{4} \operatorname{arctg}\left( \frac{e^x}{4} \right) + C \)
20
\(\displaystyle \frac{1}{2\sqrt{3}} \ln\left| \frac{\sin x - \sqrt{3}}{\sin x + \sqrt{3}} \right| + C \)
1
Facem substituția: \( u = 2x + 7 \Rightarrow du = 2dx \Rightarrow dx = \displaystyle \frac{du}{2} \)
Rezultă: \( \displaystyle \int \sqrt{2x+7} \, dx = \int \sqrt{u} \cdot \frac{1}{2} \, du = \frac{1}{2} \int u^{1/2} \, du = \frac{1}{2} \cdot \frac{u^{3/2}}{3/2} + C = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{3}(2x + 7)^{3/2} + C \)
2
Substituim: \( u = 5x - 2 \Rightarrow du = 5dx \Rightarrow dx = \displaystyle \frac{du}{5} \)
Rezultă: \( \displaystyle \int e^{u} \cdot \frac{1}{5} \, du = \frac{1}{5} \int e^u \, du = \frac{1}{5} e^{u} + C = \frac{1}{5} e^{5x - 2} + C \)
3
Substituim: \( u = 9x + 4 \Rightarrow du = 9dx \Rightarrow dx = \displaystyle \frac{du}{9} \)
Rezultă: \( \displaystyle \int \cos(u) \cdot \frac{1}{9} \, du = \frac{1}{9} \int \cos(u) \, du = \frac{1}{9} \sin(u) + C = \frac{1}{9} \sin(9x + 4) + C \)
4
Substituim: \( u = -4x + 3 \Rightarrow du = -4dx \Rightarrow dx = \displaystyle \frac{du}{-4} \)
Rezultă: \( \displaystyle \int u^{19} \cdot \frac{1}{-4} \, du = \frac{-1}{4} \cdot \frac{u^{20}}{20} + C = \frac{-1}{80}(-4x + 3)^{20} + C \)
5
Substituim: \( u = -7x + 5 \Rightarrow du = -7dx \Rightarrow dx = \displaystyle \frac{du}{-7} \)
Rezultă: \( \displaystyle \int \frac{1}{u^2} \cdot \frac{1}{-7} \, du = \frac{-1}{7} \int u^{-2} \, du = \frac{-1}{7} \cdot \frac{u^{-1}}{-1} + C = \frac{1}{7(-7x+5)} + C \)
6
Substituim: \( u = 6x - 8 \Rightarrow du = 6dx \Rightarrow dx = \displaystyle \frac{du}{6} \)
Rezultă: \( \displaystyle \int u^{-1/2} \cdot \frac{1}{6} \, du = \frac{1}{6} \cdot \frac{u^{1/2}}{1/2} + C = \frac{1}{6} \cdot 2\sqrt{u} + C = \frac{1}{3} \sqrt{6x - 8} + C \)
7
Folosim formula: \(\displaystyle \int a^{u} \, du = \frac{a^u}{\ln a} \)
Substituim: \(\displaystyle u = 5x - 3 \Rightarrow du = 5dx \Rightarrow dx = \frac{du}{5} \)
Rezultă: \(\displaystyle \int 2^u \cdot \frac{1}{5} \, du = \frac{1}{5} \cdot \frac{2^u}{\ln 2} + C = \frac{1}{5\ln 2} \cdot 2^{5x - 3} + C \)
8
Substituim: \(\displaystyle u = 8x + 1 \Rightarrow du = 8dx \Rightarrow dx = \frac{du}{8} \)
Rezultă: \(\displaystyle \int \sec^2(u) \cdot \frac{1}{8} \, du = \frac{1}{8} \operatorname{tg}(u) + C = \frac{1}{8} \operatorname{tg}(8x + 1) + C \)
9
Substituim: \(\displaystyle u = x^6 + 4 \Rightarrow du = 6x^5 dx \Rightarrow dx = \frac{du}{6x^5} \)
Rezultă: \(\displaystyle \int \frac{x^5}{\sqrt{u}} \cdot \frac{1}{6x^5} \, du = \frac{1}{6} \int u^{-1/2} \, du = \frac{1}{6} \cdot \frac{u^{1/2}}{1/2} + C = \frac{1}{3} \sqrt{x^6 + 4} + C \)
10
Substituim: \(\displaystyle u = x^2 + 3x - 1 \Rightarrow du = (2x + 3)dx \)
Rezultă: \(\displaystyle \int \frac{du}{u^2} = \int u^{-2} \, du = \frac{u^{-1}}{-1} + C = \frac{-1}{x^2 + 3x - 1} + C \)
11
Substituim: \(\displaystyle u = x^2 + 4 \Rightarrow du = 2x \, dx \Rightarrow x\,dx = \frac{du}{2} \)
Rezultă: \(\displaystyle \int \frac{1}{u} \cdot \frac{1}{2} \, du = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln(x^2 + 4) + C \)
12
Substituim: \(\displaystyle u = \sqrt{x} \Rightarrow x = u^2, \, dx = 2u \, du \)
Rezultă: \(\displaystyle \int \frac{1}{2u(u + 1)} \cdot 2u \, du = \int \frac{1}{u + 1} \, du = \ln|u + 1| + C = \ln(\sqrt{x} + 1) + C \)
13
Substituim: \(\displaystyle u = \sin x \Rightarrow du = \cos x \, dx \)
Rezultă: \(\displaystyle \int \frac{1}{u^2 + 3} \, du \)
Aceasta este de forma: \(\displaystyle \int \frac{1}{u^2 + a^2} \, du = \frac{1}{a} \operatorname{arctg}\left(\frac{u}{a}\right) + C \)
\(\displaystyle a^2 = 3 \Rightarrow a = \sqrt{3} \)
Rezultă: \(\displaystyle \frac{1}{\sqrt{3}} \operatorname{arctg}\left( \frac{\sin x}{\sqrt{3}} \right) + C \)
14
Substituim: \(\displaystyle u = 3 + x^2 \Rightarrow du = 2x \, dx \Rightarrow x\,dx = \frac{du}{2} \)
Rezultă: \(\displaystyle \int \frac{1}{\sqrt{u}} \cdot \frac{1}{2} \, du = \frac{1}{2} \int u^{-1/2} \, du = \frac{1}{2} \cdot \frac{u^{1/2}}{1/2} + C = \sqrt{3 + x^2} + C \)
15
Substituim: \(\displaystyle u = x^3 \Rightarrow du = 3x^2 dx \Rightarrow x^2 dx = \frac{du}{3} \)
Rezultă: \(\displaystyle \int e^u \cdot \frac{1}{3} \, du = \frac{1}{3} e^u + C = \frac{1}{3} e^{x^3} + C \)
16
Substituim: \(\displaystyle u = \sqrt{x} \Rightarrow x = u^2, \, dx = 2u \, du \)
Rezultă: \(\displaystyle \int \frac{7^u}{u} \cdot 2u \, du = \int 2 \cdot 7^u \, du = 2 \cdot \frac{7^u}{\ln 7} + C = \frac{2 \cdot 7^{\sqrt{x}}}{\ln 7} + C \)
17
Substituim: \(\displaystyle u = 5x + 7 \Rightarrow du = 5dx \Rightarrow dx = \frac{du}{5} \)
Rezultă: \(\displaystyle \frac{4}{3} \int u^{1/3} \cdot \frac{1}{5} \, du = \frac{4}{15} \int u^{1/3} \, du = \frac{4}{15} \cdot \frac{u^{4/3}}{4/3} + C = \frac{1}{5} u^{4/3} + C = \frac{1}{5} (5x + 7)^{4/3} + C \)
18
Este de forma \( \int \frac{1}{a^2x^2 + b^2} \, dx = \frac{1}{ab} \operatorname{arctg}\left(\frac{ax}{b}\right) + C \)
Scriem: \(\displaystyle 9x^2 + 16 = (3x)^2 + 4^2 \Rightarrow a = 3, b = 4 \)
Rezultă: \(\displaystyle \frac{1}{3 \cdot 4} \operatorname{arctg}\left( \frac{3x}{4} \right) + C = \frac{1}{12} \operatorname{arctg}\left( \frac{3x}{4} \right) + C \)
19
Substituim: \(\displaystyle u = e^x \Rightarrow du = e^x dx \)
Rezultă: \(\displaystyle \int \frac{1}{u^2 + 16} \, du = \frac{1}{4} \operatorname{arctg}\left( \frac{u}{4} \right) + C = \frac{1}{4} \operatorname{arctg}\left( \frac{e^x}{4} \right) + C \)
20
Substituim: \(\displaystyle u = \sin x \Rightarrow du = \cos x \, dx \)
Rezultă: \(\displaystyle \int \frac{1}{u^2 - 3} \, du \)
Aceasta este de forma: \(\displaystyle \int \frac{1}{u^2 - a^2} \, du = \frac{1}{2a} \ln\left|\frac{u - a}{u + a}\right| + C \)
Aici \(\displaystyle a = \sqrt{3} \Rightarrow \frac{1}{2\sqrt{3}} \ln\left| \frac{\sin x - \sqrt{3}}{\sin x + \sqrt{3}} \right| + C \)