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Substitution
Use substitution to evaluate indefinite integrals.
Proof
Let f, g, u, and F be as specified in the theorem. Then
\[\begin{array}{ll}\frac{d}{dx}F(g(x)) & ={F}^{'}\text{(}g(x)){g}^{'}\text{(}x) \\ & =f[g(x)]{g}^{'}\text{(}x).\end{array}\]Integrating both sides with respect to x, we see that
\[\int f[g(x)]{g}^{'}\text{(}x)dx=F(g(x))+C.\]If we now substitute \(u=g(x),\) and \(du=g'(x)dx,\) we get
\[\begin{array}{ll}\int f[g(x)]{g}^{'}\text{(}x)dx & =\int f(u)du \\ & =F(u)+C \\ & =F(g(x))+C.\end{array}\]□
Returning to the problem we looked at originally, we let \(u={x}^{2}-3\) and then \(du=2xdx.\) Rewrite the integral in terms of u:
\[\int {\underset{u}{\underset{⏟}{({x}^{2}-3)}}}^{3}\underset{du}{\underset{⏟}{(2xdx)}}=\int {u}^{3}du.\]Using the power rule for integrals, we have
\[\int {u}^{3}du=\frac{{u}^{4}}{4}+C.\]Substitute the original expression for x back into the solution:
\[\frac{{u}^{4}}{4}+C=\frac{{({x}^{2}-3)}^{4}}{4}+C.\]Example
Try it.
Use substitution to find the antiderivative \(\int 6x{(3{x}^{2}+4)}^{4}dx.\)
Solution
The first step is to choose an expression for u. We choose \(u=3{x}^{2}+4\) because then \(du=6xdx,\) and we already have du in the integrand. Write the integral in terms of u:
\[\int 6x{(3{x}^{2}+4)}^{4}dx=\int {u}^{4}du.\]Remember that du is the derivative of the expression chosen for u, regardless of what is inside the integrand. Now we can evaluate the integral with respect to u:
\[\begin{array}{ll}\int {u}^{4}du & =\frac{{u}^{5}}{5}+C \\ \\ \\ & =\frac{{(3{x}^{2}+4)}^{5}}{5}+C.\end{array}\]Condensed: the full section is in OpenStax Calculus Volume 2.
Substitution for Definite Integrals
Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.
Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if \(F(x)\) is an antiderivative of \(f(x),\) we have
\[\int f(g(x)){g}^{'}\text{(}x)dx=F(g(x))+C.\]Then
\[\begin{array}{ll}{\int }_{a}^{b}f[g(x)]{g}^{'}\text{(}x)dx & ={F(g(x))|}_{x=a}^{x=b} \\ & =F(g(b))-F(g(a)) \\ & ={F(u)|}_{u=g(a)}^{u=g(b)} \\ \\ \\ & ={\int }_{g(a)}^{g(b)}f(u)du,\end{array}\]and we have the desired result.
Example
Try it.
Use substitution to evaluate \({\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx.\)
Solution
Let \(u=1+2{x}^{3},\) so \(du=6{x}^{2}dx.\) Since the original function includes one factor of x2 and \(du=6{x}^{2}dx,\) multiply both sides of the du equation by \(1\text{/}6.\) Then,
\[\begin{array}{lll}du & = & 6{x}^{2}dx \\ \frac{1}{6}du & = & {x}^{2}dx.\end{array}\]To adjust the limits of integration, note that when \(x=0,u=1+2(0)=1,\) and when \(x=1,u=1+2(1)=3.\) Then
\[{\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx=\frac{1}{6}{\int }_{1}^{3}{u}^{5}du.\]Evaluating this expression, we get
\[\begin{array}{ll} \\ \\ \frac{1}{6}{\int }_{1}^{3}{u}^{5}du & =(\frac{1}{6})(\frac{{u}^{6}}{6}){|}_{1}^{3} \\ & =\frac{1}{36}[{(3)}^{6}-{(1)}^{6}] \\ & =\frac{182}{9}.\end{array}\]Example
Try it.
Use substitution to evaluate \({\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx.\)
Solution
Let \(u=4{x}^{2}+3.\) Then, \(du=8xdx.\) To adjust the limits of integration, we note that when \(x=0,u=3,\) and when \(x=1,u=7.\) So our substitution gives
\[\begin{array}{ll}{\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx & =\frac{1}{8}{\int }_{3}^{7}{e}^{u}du \\ \\ & =\frac{1}{8}{e}^{u}{|}_{3}^{7} \\ & =\frac{{e}^{7}-{e}^{3}}{8} \\ & \approx 134.568.\end{array}\]Condensed: the full section is in OpenStax Calculus Volume 2.
Key Concepts
- Substitution is a technique that simplifies the integration of functions that are the result of a chain-rule derivative. The term ‘substitution’ refers to changing variables or substituting the variable u and du for appropriate expressions in the integrand.
- When using substitution for a definite integral, we also have to change the limits of integration.
Key Equations
| Substitution with Indefinite Integrals | \(\int f[g(x)]{g}^{'}\text{(}x)dx=\int f(u)du=F(u)+C=F(g(x))+C\) |
| Substitution with Definite Integrals | \({\int }_{a}^{b}f(g(x))g'(x)dx={\int }_{g(a)}^{g(b)}f(u)du\) |
Substitution
In the following exercises, verify each identity using differentiation. Then, using the indicated u-substitution, identify f such that the integral takes the form \(\int f(u)du.\)
In the following exercises, find the antiderivative using the indicated substitution.
In the following exercises, use a suitable change of variables to determine the indefinite integral.
In the following exercises, use a calculator to estimate the area under the curve using left Riemann sums with 50 terms, then use substitution to solve for the exact answer.
In the following exercises, use a change of variables to evaluate the definite integral.
In the following exercises, evaluate the indefinite integral \(\int f(x)dx\) with constant \(C=0\) using u-substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of C that would need to be added to the antiderivative to make it equal to the definite integral \(F(x)={\int }_{a}^{x}f(t)dt,\) with a the left endpoint of the given interval.
In the following exercises, use a change of variables to show that each definite integral is equal to zero.
Proof
Let f, g, u, and F be as specified in the theorem. Then
\[\begin{array}{ll}\frac{d}{dx}F(g(x)) & ={F}^{'}\text{(}g(x)){g}^{'}\text{(}x) \\ & =f[g(x)]{g}^{'}\text{(}x).\end{array}\]Integrating both sides with respect to x, we see that
\[\int f[g(x)]{g}^{'}\text{(}x)dx=F(g(x))+C.\]If we now substitute \(u=g(x),\) and \(du=g'(x)dx,\) we get
\[\begin{array}{ll}\int f[g(x)]{g}^{'}\text{(}x)dx & =\int f(u)du \\ & =F(u)+C \\ & =F(g(x))+C.\end{array}\]□
Returning to the problem we looked at originally, we let \(u={x}^{2}-3\) and then \(du=2xdx.\) Rewrite the integral in terms of u:
\[\int {\underset{u}{\underset{⏟}{({x}^{2}-3)}}}^{3}\underset{du}{\underset{⏟}{(2xdx)}}=\int {u}^{3}du.\]Using the power rule for integrals, we have
\[\int {u}^{3}du=\frac{{u}^{4}}{4}+C.\]Substitute the original expression for x back into the solution:
\[\frac{{u}^{4}}{4}+C=\frac{{({x}^{2}-3)}^{4}}{4}+C.\]Example
Try it.
Use substitution to find the antiderivative \(\int 6x{(3{x}^{2}+4)}^{4}dx.\)
Solution
The first step is to choose an expression for u. We choose \(u=3{x}^{2}+4\) because then \(du=6xdx,\) and we already have du in the integrand. Write the integral in terms of u:
\[\int 6x{(3{x}^{2}+4)}^{4}dx=\int {u}^{4}du.\]Remember that du is the derivative of the expression chosen for u, regardless of what is inside the integrand. Now we can evaluate the integral with respect to u:
\[\begin{array}{ll}\int {u}^{4}du & =\frac{{u}^{5}}{5}+C \\ \\ \\ & =\frac{{(3{x}^{2}+4)}^{5}}{5}+C.\end{array}\]Condensed: the full section is in OpenStax Calculus Volume 1.
Substitution for Definite Integrals
Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.
Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if \(F(x)\) is an antiderivative of \(f(x),\) we have
\[\int f(g(x)){g}^{'}\text{(}x)dx=F(g(x))+C.\]Then
\[\begin{array}{ll}{\int }_{a}^{b}f[g(x)]{g}^{'}\text{(}x)dx & ={F(g(x))|}_{x=a}^{x=b} \\ & =F(g(b))-F(g(a)) \\ & ={F(u)|}_{u=g(a)}^{u=g(b)} \\ \\ \\ & ={\int }_{g(a)}^{g(b)}f(u)du,\end{array}\]and we have the desired result.
Example
Try it.
Use substitution to evaluate \({\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx.\)
Solution
Let \(u=1+2{x}^{3},\) so \(du=6{x}^{2}dx.\) Since the original function includes one factor of x2 and \(du=6{x}^{2}dx,\) multiply both sides of the du equation by \(1\text{/}6.\) Then,
\[\begin{array}{lll}du & = & 6{x}^{2}dx \\ \frac{1}{6}du & = & {x}^{2}dx.\end{array}\]To adjust the limits of integration, note that when \(x=0,u=1+2(0)=1,\) and when \(x=1,u=1+2(1)=3.\) Then
\[{\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx=\frac{1}{6}{\int }_{1}^{3}{u}^{5}du.\]Evaluating this expression, we get
\[\begin{array}{ll} \\ \\ \frac{1}{6}{\int }_{1}^{3}{u}^{5}du & =(\frac{1}{6})(\frac{{u}^{6}}{6}){|}_{1}^{3} \\ & =\frac{1}{36}[{(3)}^{6}-{(1)}^{6}] \\ & =\frac{182}{9}.\end{array}\]Example
Try it.
Use substitution to evaluate \({\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx.\)
Solution
Let \(u=4{x}^{2}+3.\) Then, \(du=8xdx.\) To adjust the limits of integration, we note that when \(x=0,u=3,\) and when \(x=1,u=7.\) So our substitution gives
\[\begin{array}{ll}{\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx & =\frac{1}{8}{\int }_{3}^{7}{e}^{u}du \\ \\ & =\frac{1}{8}{e}^{u}{|}_{3}^{7} \\ & =\frac{{e}^{7}-{e}^{3}}{8} \\ & \approx 134.568.\end{array}\]Condensed: the full section is in OpenStax Calculus Volume 1.
Key Concepts
- Substitution is a technique that simplifies the integration of functions that are the result of a chain-rule derivative. The term ‘substitution’ refers to changing variables or substituting the variable u and du for appropriate expressions in the integrand.
- When using substitution for a definite integral, we also have to change the limits of integration.
Key Equations
| Substitution with Indefinite Integrals | \(\int f[g(x)]{g}^{'}\text{(}x)dx=\int f(u)du=F(u)+C=F(g(x))+C\) |
| Substitution with Definite Integrals | \({\int }_{a}^{b}f(g(x))g'(x)dx={\int }_{g(a)}^{g(b)}f(u)du\) |
Substitution
In the following exercises, verify each identity using differentiation. Then, using the indicated u-substitution, identify f such that the integral takes the form \(\int f(u)du.\)
In the following exercises, find the antiderivative using the indicated substitution.
In the following exercises, use a suitable change of variables to determine the indefinite integral.
In the following exercises, use a calculator to estimate the area under the curve using left Riemann sums with 50 terms, then use substitution to solve for the exact answer.
In the following exercises, use a change of variables to evaluate the definite integral.
In the following exercises, evaluate the indefinite integral \(\int f(x)dx\) with constant \(C=0\) using u-substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of C that would need to be added to the antiderivative to make it equal to the definite integral \(F(x)={\int }_{a}^{x}f(t)dt,\) with a the left endpoint of the given interval.
In the following exercises, use a change of variables to show that each definite integral is equal to zero.
İndi siz Heç bir hesablama proqramı bu ədədi həll edə bilməz, lakin onun hissələri hesablana bilər. Aşağıdakılardan birini sınayın ya da özünüz yazın.
Tədris (40)
Hər bir cavabı əvvəlcə kağız üzərində sınayın. Cavabı yoxlamaq üçün göstərin; təsdiqlənmiş cavablar hər addım üçün həlledicidə açıla bilər.
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Use substitution to find the antiderivative \(\int 6x{(3{x}^{2}+4)}^{4}dx.\)
Cavabı göstər
The first step is to choose an expression for u. We choose \(u=3{x}^{2}+4\) because then \(du=6xdx,\) and we already have du in the integrand. Write the integral in terms of u:
\[\int 6x{(3{x}^{2}+4)}^{4}dx=\int {u}^{4}du.\]Remember that du is the derivative of the expression chosen for u, regardless of what is inside the integrand. Now we can evaluate the integral with respect to u:
\[\begin{array}{ll}\int {u}^{4}du & =\frac{{u}^{5}}{5}+C \\ \\ \\ & =\frac{{(3{x}^{2}+4)}^{5}}{5}+C.\end{array}\] -
Use substitution to find the antiderivative \(\int 3{x}^{2}{({x}^{3}-3)}^{2}dx.\)
Cavabı göstər
\(\int 3{x}^{2}{({x}^{3}-3)}^{2}dx=\frac{1}{3}{({x}^{3}-3)}^{3}+C\)
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Use substitution to find \(\int z\sqrt{{z}^{2}-5}dz.\)
Cavabı göstər
Rewrite the integral as \(\int z{({z}^{2}-5)}^{1\text{/}2}dz.\) Let \(u={z}^{2}-5\) and \(du=2z\ dz.\) Now we have a problem because \(du=2z\ dz\) and the original expression has only \(z\ dz.\) We have to alter our expression for du or the integral in u will be twice as large as it should be. If we multiply both sides of the du equation by \(\frac{1}{2}.\) we can solve this problem. Thus,
\[\begin{array}{ll} \\ u & ={z}^{2}-5 \\ du & =2z\ dz \\ \frac{1}{2}du & =\frac{1}{2}(2z)dz=z\ dz.\end{array}\]Write the integral in terms of u, but pull the \(\frac{1}{2}\) outside the integration symbol:
\[\int z{({z}^{2}-5)}^{1\text{/}2}dz=\frac{1}{2}\int {u}^{1\text{/}2}du.\]Integrate the expression in u:
\[\begin{array}{ll} \\ \frac{1}{2}\int {u}^{1\text{/}2}du & =(\frac{1}{2})\frac{{u}^{3\text{/}2}}{\frac{3}{2}}+C \\ \\ & =(\frac{1}{2})(\frac{2}{3}){u}^{3\text{/}2}+C \\ & =\frac{1}{3}{u}^{3\text{/}2}+C \\ & =\frac{1}{3}{({z}^{2}-5)}^{3\text{/}2}+C.\end{array}\] -
Use substitution to find \(\int {x}^{2}{({x}^{3}+5)}^{9}dx.\)
Cavabı göstər
\(\frac{{({x}^{3}+5)}^{10}}{30}+C\)
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Use substitution to evaluate the integral \(\int \frac{\text{sin}\ t}{{\text{cos}}^{3}t}dt.\)
Cavabı göstər
We know the derivative of \(\text{cos}\ t\) is \(\text{-}\text{sin}\ t,\) so we set \(u=\text{cos}\ t.\) Then \(du=\text{-}\text{sin}\ tdt.\) Substituting into the integral, we have
\[\int \frac{\text{sin}\ t}{{\text{cos}}^{3}t}dt=\text{-}\int \frac{du}{{u}^{3}}.\]Evaluating the integral, we get
\[\begin{array}{ll} \\ \\ \text{-}\int \frac{du}{{u}^{3}} & =\text{-}\int {u}^{-3}du \\ & =\text{-}(-\frac{1}{2}){u}^{-2}+C.\end{array}\]Putting the answer back in terms of t, we get
\[\begin{array}{ll}\int \frac{\text{sin}\ t}{{\text{cos}}^{3}t}dt & =\frac{1}{2{u}^{2}}+C \\ \\ & =\frac{1}{2{\text{cos}}^{2}t}+C.\end{array}\] -
Use substitution to evaluate the integral \(\int \frac{\text{cos}\ t}{{\text{sin}}^{2}t}dt.\)
Cavabı göstər
\(-\frac{1}{\text{sin}\ t}+C\)
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Use substitution to find the antiderivative \(\int \frac{x}{\sqrt{x-1}}dx.\)
Cavabı göstər
If we let \(u=x-1,\) then \(du=dx.\) But this does not account for the x in the numerator of the integrand. We need to express x in terms of u. If \(u=x-1,\) then \(x=u+1.\) Now we can rewrite the integral in terms of u:
\[\begin{array}{ll}\int \frac{x}{\sqrt{x-1}}dx & =\int \frac{u+1}{\sqrt{u}}du \\ \\ & =\int \sqrt{u}+\frac{1}{\sqrt{u}}du \\ & =\int ({u}^{1\text{/}2}+{u}^{-1\text{/}2})du.\end{array}\]Then we integrate in the usual way, replace u with the original expression, and factor and simplify the result. Thus,
\[\begin{array}{ll}\int ({u}^{1\text{/}2}+{u}^{-1\text{/}2})du & =\frac{2}{3}{u}^{3\text{/}2}+2{u}^{1\text{/}2}+C \\ \\ & =\frac{2}{3}{(x-1)}^{3\text{/}2}+2{(x-1)}^{1\text{/}2}+C \\ & ={(x-1)}^{1\text{/}2}[\frac{2}{3}(x-1)+2]+C \\ & ={(x-1)}^{1\text{/}2}(\frac{2}{3}x-\frac{2}{3}+\frac{6}{3})+C \\ & ={(x-1)}^{1\text{/}2}(\frac{2}{3}x+\frac{4}{3})+C \\ & =\frac{2}{3}{(x-1)}^{1\text{/}2}(x+2)+C.\end{array}\] -
Use substitution to evaluate the indefinite integral \(\int {\text{cos}}^{3}t\ \text{sin}\ t\ dt.\)
Cavabı göstər
\(-\frac{{\text{cos}}^{4}t}{4}+C\)
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Use substitution to evaluate \({\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx.\)
Cavabı göstər
Let \(u=1+2{x}^{3},\) so \(du=6{x}^{2}dx.\) Since the original function includes one factor of x2 and \(du=6{x}^{2}dx,\) multiply both sides of the du equation by \(1\text{/}6.\) Then,
\[\begin{array}{lll}du & = & 6{x}^{2}dx \\ \frac{1}{6}du & = & {x}^{2}dx.\end{array}\]To adjust the limits of integration, note that when \(x=0,u=1+2(0)=1,\) and when \(x=1,u=1+2(1)=3.\) Then
\[{\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx=\frac{1}{6}{\int }_{1}^{3}{u}^{5}du.\]Evaluating this expression, we get
\[\begin{array}{ll} \\ \\ \frac{1}{6}{\int }_{1}^{3}{u}^{5}du & =(\frac{1}{6})(\frac{{u}^{6}}{6}){|}_{1}^{3} \\ & =\frac{1}{36}[{(3)}^{6}-{(1)}^{6}] \\ & =\frac{182}{9}.\end{array}\] -
Use substitution to evaluate the definite integral \({\int }_{-1}^{0}y{(2{y}^{2}-3)}^{5}dy.\)
Cavabı göstər
\(\frac{91}{3}\)
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Use substitution to evaluate \({\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx.\)
Cavabı göstər
Let \(u=4{x}^{2}+3.\) Then, \(du=8xdx.\) To adjust the limits of integration, we note that when \(x=0,u=3,\) and when \(x=1,u=7.\) So our substitution gives
\[\begin{array}{ll}{\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx & =\frac{1}{8}{\int }_{3}^{7}{e}^{u}du \\ \\ & =\frac{1}{8}{e}^{u}{|}_{3}^{7} \\ & =\frac{{e}^{7}-{e}^{3}}{8} \\ & \approx 134.568.\end{array}\] -
Use substitution to evaluate \({\int }_{0}^{1}{x}^{2}\text{cos}(\frac{\pi }{2}{x}^{3})dx.\)
Cavabı göstər
\(\frac{2}{3\pi }\approx 0.2122\)
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Use substitution to evaluate \({\int }_{0}^{\pi \text{/}2}{\text{cos}}^{2}\theta \ d\theta .\)
Cavabı göstər
Let us first use a trigonometric identity to rewrite the integral. The trig identity \({\text{cos}}^{2}\theta =\frac{1+\text{cos}\ 2\theta }{2}\) allows us to rewrite the integral as
\[{\int }_{0}^{\pi \text{/}2}{\text{cos}}^{2}\theta d\theta ={\int }_{0}^{\pi \text{/}2}\frac{1+\text{cos}\ 2\theta }{2}d\theta .\]Then,
\[\begin{array}{ll}{\int }_{0}^{\pi \text{/}2}(\frac{1+\text{cos}\ 2\theta }{2})d\theta & ={\int }_{0}^{\pi \text{/}2}(\frac{1}{2}+\frac{1}{2}\text{cos}\ 2\theta )d\theta \\ \\ \\ & =\frac{1}{2}{\int }_{0}^{\pi \text{/}2}d\theta +\frac{1}{2}{\int }_{0}^{\pi \text{/}2}\text{cos}\ 2\theta d\theta .\end{array}\]We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let \(u=2\theta .\) Then, \(du=2d\theta ,\) or \(\frac{1}{2}du=d\theta .\) Also, when \(\theta =0,u=0,\) and when \(\theta =\pi \text{/}2,u=\pi .\) Expressing the second integral in terms of u, we have
\[\begin{array}{ll} \\ \\ \frac{1}{2}{\int }_{0}^{\pi \text{/}2}d\theta +\frac{1}{2}{\int }_{0}^{\pi \text{/}2}\text{cos}\ 2\theta d\theta & =\frac{1}{2}{\int }_{0}^{\pi \text{/}2}d\theta +\frac{1}{2}(\frac{1}{2}){\int }_{0}^{\pi }\text{cos}\ udu \\ & =\frac{\theta }{2}{|}_{\theta =0}^{\theta =\pi \text{/}2}+\frac{1}{4}\text{sin}\ u{|}_{u=0}^{u=\pi } \\ & =(\frac{\pi }{4}-0)+(0-0)=\frac{\pi }{4}.\end{array}\] -
Why is u-substitution referred to as change of variable?
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2. If \(f=g∘h,\) when reversing the chain rule, \(\frac{d}{dx}(g∘h)(x)={g}^{'}\text{(}h(x)){h}^{'}\text{(}x),\) should you take \(u=g(x)\) or \(u=h(x)?\)
Cavabı göstər
\(u=h(x)\)
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\(\int x\sqrt{x+1}dx=\frac{2}{15}{(x+1)}^{3\text{/}2}(3x-2)+C;u=x+1\)
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For \(x>1:\int \frac{{x}^{2}}{\sqrt{x-1}}dx=\frac{2}{15}\sqrt{x-1}(3{x}^{2}+4x+8)+C;u=x-1\)
Cavabı göstər
\(f(u)=\frac{{(u+1)}^{2}}{\sqrt{u}}\)
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\(\int x\sqrt{4{x}^{2}+9}dx=\frac{1}{12}{(4{x}^{2}+9)}^{3\text{/}2}+C;u=4{x}^{2}+9\)
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\(\int \frac{x}{\sqrt{4{x}^{2}+9}}dx=\frac{1}{4}\sqrt{4{x}^{2}+9}+C;u=4{x}^{2}+9\)
Cavabı göstər
\(du=8xdx;f(u)=\frac{1}{8\sqrt{u}}\)
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\(\int \frac{x}{{(4{x}^{2}+9)}^{2}}dx=-\frac{1}{8(4{x}^{2}+9)};u=4{x}^{2}+9\)
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\(\int {(x+1)}^{4}dx;u=x+1\)
Cavabı göstər
\(\frac{1}{5}{(x+1)}^{5}+C\)
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\(\int {(x-1)}^{5}dx;u=x-1\)
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\(\int {(2x-3)}^{-7}dx;u=2x-3\)
Cavabı göstər
\(-\frac{1}{12{(2x-3)}^{6}}+C\)
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\(\int {(3x-2)}^{-11}dx;u=3x-2\)
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\(\int \frac{x}{\sqrt{{x}^{2}+1}}dx;u={x}^{2}+1\)
Cavabı göstər
\(\sqrt{{x}^{2}+1}+C\)
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\(\int \frac{x}{\sqrt{1-{x}^{2}}}dx;u=1-{x}^{2}\)
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\(\int (x-1){({x}^{2}-2x)}^{3}dx;u={x}^{2}-2x\)
Cavabı göstər
\(\frac{1}{8}{({x}^{2}-2x)}^{4}+C\)
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\(\int ({x}^{2}-2x){({x}^{3}-3{x}^{2})}^{2}dx;u={x}^{3}-3{x}^{2}\)
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\(\int {\text{cos}}^{3}\theta d\theta ;u=\text{sin}\ \theta\) \(\text{(}Hint\text{:}\ {\text{cos}}^{2}\theta =1-{\text{sin}}^{2}\theta \text{)}\)
Cavabı göstər
\(\text{sin}\ \theta -\frac{{\text{sin}}^{3}\theta }{3}+C\)
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\(\int {\text{sin}}^{3}\theta d\theta ;u=\text{cos}\ \theta\) \(\text{(}Hint\text{:}\ {\text{sin}}^{2}\theta =1-{\text{cos}}^{2}\theta \text{)}\)
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\(\int x{(1-x)}^{99}dx\)
Cavabı göstər
\(\frac{{(1-x)}^{101}}{101}-\frac{{(1-x)}^{100}}{100}+C\)
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\(\int t{(1-{t}^{2})}^{10}dt\)
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\(\int {(11x-7)}^{-3}dx\)
Cavabı göstər
\(\int {(11x-7)}^{-3}\text{dx}=-\frac{1}{22{(11x-7)}^{2}}+C\)
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\(\int {(7x-11)}^{4}dx\)
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\(\int {\text{cos}}^{3}\theta \ \text{sin}\ \theta d\theta\)
Cavabı göstər
\(-\frac{{\text{cos}}^{4}\theta }{4}+C\)
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\(\int {\text{sin}}^{7}\theta \ \text{cos}\ \theta d\theta\)
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\(\int {\text{cos}}^{2}(\pi t)\text{sin}(\pi t)dt\)
Cavabı göstər
\(-\frac{{\text{cos}}^{3}(\pi t)}{3\pi }+C\)
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\(\int {\text{sin}}^{2}x{\text{cos}}^{3}xdx\) \(\text{(}Hint\text{:}\ {\text{sin}}^{2}x+{\text{cos}}^{2}x=1\text{)}\)
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\(\int t\ \text{sin}({t}^{2})\text{cos}({t}^{2})dt\)
Cavabı göstər
\(-\frac{1}{4}\ {\text{cos}}^{2}({t}^{2})+C\)
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\(\int {t}^{2}{\text{cos}}^{2}({t}^{3})\text{sin}({t}^{3})dt\)
Burada istifadə edilən simvollar
Tam təsviri, şəkli və hər hərfin nə demək olduğunu görmək üçün hər hansı simvolu vurun.
Növbəti: Substitution
- Use substitution to evaluate indefinite integrals.
- Use substitution to evaluate definite integrals.
- Look carefully at the integrand and select an expression
- Substitute
- Evaluate the integral in terms of
- Write the result in terms of
- Substitution is a technique that simplifies the integration of functions that are the result of a chain-rule derivative. The term ‘substitution’ refers to changing variables or substituting the variable
- When using substitution for a definite integral, we also have to change the limits of integration.
İnsanlar soruşurlar
What is a derivative in one sentence?
The slope of the graph at a point, the rate at which the output is changing there. Speed is the derivative of position.
What is an integral in one sentence?
The accumulated total of a rate: the area under the curve. Distance travelled is the integral of speed.
Why are derivatives and integrals opposites?
That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.
When do I use substitution and when integration by parts?
Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function, a polynomial times an exponential, log or trig function.
Bu səhifənin bəzi hissələri OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Bu, bizim üçün çox böyük bir uğurdur, biz bunu çox istəyirik. Onun mənbəyi kommersiya məqsədləri üçün lisenziyalı olduğundan, bu dərs pulsuzdur, heç nə satmır və eyni lisenziyaya əsasən paylaşılır.
Daha çox Calculus
LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests