Що таке u-заміна?

In this section, we'll learn about another technique for integration, called $u$-substitution. But before we start, let's look at a problem:

The first two choices are technically possible ways to solve this integral, but a substitution seems to be the simplest option. Recall that we have the power rule $$\int x^n\, dx=\frac{1}{n+1}x^{n+1}+C.$$ So, for instance, we know that $$\int u^7\, du=\frac{1}{8}u^8+C.$$ So what happens if we set $u=x+1$? Well, that gives us $$\frac{du}{dx}=1\Longrightarrow du=dx.$$ Now, we can substitute these expressions into the initial integral to find: $$\int (x+1)^7\, dx=\int u^7\, du=\frac{1}{8}u^8+C.$$ Now, remember that $u=x+1$. So, our final answer becomes $$\int (x+1)^7\, dx=\frac{1}{8}u^8+C=\frac{1}{8}(x+1)^8+C.$$ Wow, that was a lot simpler than expected! This is the power of the technique called $u$-substitution, in which we substitute a section of the integrand with a new variable to simplify the integral. Let's look at another example together, and evaluate the integral $$\int x\sqrt{x^2+1}\, dx.$$ First, try and think about what we should substitute for $u$.

In this problem, the most convenient choice for $u$ would be $u=x^2+1$. This gives us $$\frac{du}{dx}=2x\Longrightarrow \frac{1}{2}\, du=x\, dx.$$ Now, we can return to the initial integral: $$\int x\sqrt{x^2+1}\, dx=\int \sqrt{x^2+1}\cdot (x\, dx)=\int \sqrt{u}\cdot \left(\frac{1}{2}\, du\right)$$ $$=\frac{1}{2}\int \sqrt{u}\, du=\frac{1}{2}\left(\frac{2}{3}u^{\frac{3}{2}}+c\right)=\frac{1}{3}(x^2+1)^{\frac{3}{2}}+C,$$ where $C=\frac{1}{2}c$ is just a new constant.

Did you notice a pattern for how we choose our expressions for $u$? We try to set $u=f(x)$ such that $f'(x)$ is also a part of the integrand. This is because $$\frac{du}{dx}=f'(x)\Longrightarrow du=f'(x)\, dx,$$ and so we need the factor of $f'(x)$ to replace with $du$. Hence, in the previous example we set $u=x^2+1$ because $du=2x$ was a part of the integral.

Now you try!

As before, we notice that $$\frac{d}{dx} (x^3+2)=3x^2,$$ which is also a part of the integrand! So, we set $u=x^3+2$. Then $$\frac{du}{dx}=3x^2\Longrightarrow du=3x^2\, dx,$$ and so the integral becomes $$\int 3x^2\cos{(x^3+2)}\, dx=\int \cos{(x^3+2)}\cdot (3x^2\, dx)=\int \cos{u}\, du$$ $$=\sin{u}+C=\sin{(x^3+2)}+C.$$

So far, we've only been dealing with substitutions of polynomials, but this isn't always the case! Let's now try to evaluate $$\int \tan{x}\, dx.$$ Wait a moment! There aren't even two parts to this function! Well, we can rewrite this as $$\int \frac{\sin{x}}{\cos{x}}\, dx.$$ Aha! Now we see two different parts. Moreover, we also recall that $$\frac{d}{dx} (\cos{x})=-\sin{x},$$ which is part of the integral! So, we can set $u=\cos{x}$, which gives us $$\frac{du}{dx}=-\sin{x}\, dx\Longrightarrow -du=\sin{x}\, dx.$$ Therefore, our original integral becomes $$\int \frac{\sin{x}}{\cos{x}}\, dx=\int \frac{1}{\cos{x}}\cdot(\sin{x}\, dx)=\int \frac{1}{u}\cdot (-du)=-\int \frac{du}{u}$$ $$=-\ln{|u|}+C=-\ln{|\cos{u}|}+C.$$

Try the next one on your own!

Let's approach this integral from three different ways:

1. Let's use $u$-substitution, and let $u=\sin{x}$. Then $$\frac{du}{dx}=\cos{x}\Longrightarrow du=\cos{x}\, dx.$$ So, the integral becomes $$\int \sin{x}\cos{x}\, dx=\int u\, du=\frac{u^2}{2}+C=\frac{\sin^2{x}}{2}+C.$$

2. Let's again use $u$-substitution, but this time let $u=\cos{x}$. Then $$\frac{du}{dx}=-\sin{x}\Longrightarrow du=-\sin{x}\, dx.$$ So, the integral becomes $$\int \sin{x}\cos{x}\, dx=-\int \cos{x}\cdot(-\sin{x}\, dx)\\=-\int u\, du=-\frac{u^2}{2}+C=-\frac{\cos^2{x}}{2}+C.$$

3. Lastly, let's try and use a clever trick: remember the trigonometric identity $$\sin{(2x)}=2\sin{x}\cos{x}\Longrightarrow \frac{\sin{(2x)}}{2}=\sin{x}\cos{x}.$$ So, the integral becomes $$\int \sin{x}\cos{x}\, dx=\int \frac{1}{2}\sin{(2x)}\, dx=-\frac{1}{4}\cos{(2x)}+C.$$

Hm... All of our methods seem valid, but the expressions $$\frac{\sin^2{x}}{2}+C, -\frac{\cos^2{x}}{2}+C, -\frac{\cos{(2x)}}{4}+C$$ all look different!

Well actually, the trick is that these functions are all the same in disguise! Try and figure it out!

First, remember the Pythagorean Identity: $$\sin^2{x}+\cos^2{x}=1.$$ So, if we look at the first expression, we see that $$\frac{\sin^2{x}}{2}+C=\frac{1-\cos^2{x}}{2}+C=-\frac{\cos^2{x}}{2}+\left(\frac{1}{2}+C\right),$$ which is exactly the second expression above, with just a shifted constant term.

Similarly, recall the cosine double angle formula: $$\cos{(2x)}=2\cos^2{x}-1\Longrightarrow \cos^2{x}=\frac{\cos{(2x)}+1}{2}.$$ So, the second expression now becomes $$-\frac{\cos^2{x}}{2}+C=-\frac{\frac{\cos{(2x)}+1}{2}}{2}+C\\=-\frac{\cos{(2x)}+1}{4}+C=-\frac{\cos{(2x)}}{2}+\left(-\frac{1}{4}+C\right),$$ which is exactly the third expression above, with another shifted constant term. So, these three answers were actually the same! Phew!

Now, up to this point we've only been talking about $u$-substitution with indefinite integrals. But actually, this technique also works when we want to evaluate definite integrals. (This should sound reasonable, since when we solve definite integrals we simply find the indefinite integral and plug in the limits of integration.)

So, let's say we want to figure out the value of $$\int_0^{\frac{\pi}{2}} \sin{x}\cos{x}\, dx$$ Using the substitution $u=\sin{x}$, we found earlier that $$\int \sin{x}\cos{x}\, dx=\frac{\sin^2{x}}{2}+C.$$ So, it follows that $$\int_0^{\frac{\pi}{2}} \sin{x}\cos{x}\, dx=\frac{\sin^2{x}}{2}\bigg|_0^{\frac{\pi}{2}}=\frac{1^2}{2}-\frac{0^2}{2}=\frac{1}{2}.$$ Alternatively, using $u$-substitution we had $$\int \sin{x}\cos{x}\, x=\frac{u^2}{2}+C.$$ Since $u(0)=\sin{0}=0$ and $u\left(\frac{\pi}{2}\right)=\sin{\left(\frac{\pi}{2}\right)}=1$, we can instead solve the integral as $$\int_0^{\frac{\pi}{2}} \sin{x}\cos{x}\, x=\frac{u^2}{2}\bigg|_0^1=\frac{1^2}{2}-\frac{0^2}{2}=\frac{1}{2}.$$

Try another example!

Let's go through this together: previously, we discovered that $$\int x\sqrt{x^2+1}=\frac{1}{3}(x^2+1)^{\frac{3}{2}}+C.$$ So, the definite integral becomes $$\int_0^1 x\sqrt{x^2+1}=\frac{1}{3}(x^2+1)^{\frac{3}{2}}\bigg|_0^1=\frac{1}{3}(1^2+1)^{\frac{3}{2}}-\frac{1}{3}(0^2+1)^{\frac{3}{2}}=\frac{1}{3}(2\sqrt{2}-1).$$ Alternatively, recall that during the $u$-substitution we set $u=x^2+1$, which then implied that $$\int x\sqrt{x^2+1}=\frac{1}{3}u^{\frac{3}{2}}+C.$$ So, since $u(0)=0^2+1=1$ and $u(1)=1^2+1=2$, we have $$\int_0^1 x\sqrt{x^2+1}=\frac{1}{3}u^{\frac{3}{2}}\bigg|1^2=\frac{1}{3}(2\sqrt{2}-1).$$

As you saw in this section, $u$-substitution can be very useful in integrals. However, this isn't the only kind of $u$-substitution! Click on the next section to learn about another flavor of this technique!

We can take the derivative of a function by using the Chain Rule $f(g(x))=f'(g(x))g'(x)$. The find the anti-derivative of a function with an "outside" and an "inside," we use U-substitution.

We set the inside function equal to $u$ and then find $du$.

Next, we take the anti-derivative of the simplified function with respect to $u$.

Then, we put the inside function back into the new function and $u$ disappears.

Finally, we have our answer.

Now, let's do an example: Let's say we have $\int_ {}5\cos(5x)) dx$. The outside function is $5\cos(x)$ and the inside function is $5x$. We set $u=5x$ and therefore $du=5dx$. Now, we have $\int_ {}\cos(u) du$. (The 5 out front disappeared because $du=5dx$) Taking the integral now, we have $\sin(u)$ and then if we put back $5x$ for $u$, we have our final answer, $f(g(x))=\sin(5x)$ If we want to check our answer, we can just take the derivative of this function using the chain rule.

Sometimes there are functions which cannot be integrated easily.

$\int_0^1 (3t-1)^5 dx$

The integral above, for example, cannot be integrated in the current form.

However, an integral that looks like this $\int_0^1 (x)^5 dx$ can be integrated algebraically in a way that $\int_0^1 (3t-1)^5 dx$ cannot be.

In order to integrate $\int_0^1 (3t-1)^5 dx$ the form of the integral must be changed into something more similar to $\int_0^1 (x)^5 dx$.

To do this, we use the substitution rule, which is that $\int(f(g(x)))g'(x) dx$=$\int(f(u))du$.

This is called u-substitution.

In order to integrate our original problem, we have to let $u=3t-1$ and $du=3 \, dx$, so

$\int_0^1 (3t-1)^5 dx=\frac13 \int_0^1 (u)^5 du$

Because $du=3\, dx$, the integral must be divided by three. This is a much easier integral to solve than the initial problem.

$\int_0^1 (3t-1)^5 \, dx = \frac13 \left. \dfrac{u^6}{6}\right|_0^1=\frac13\left(\dfrac{(0)^6}{6} - \dfrac{(1)^6}{6}\right)=\frac{1}{18}.$

(Video) Integration Using U-Substitution

by PatrickJMT

In this video by PatrickJMT he teaches how to use u-substitution in order to solve tricky integrals.

U-substitution is a method used to turn harder to solve integrals into a more recognizable one. Let's look at and example to see. $$\int 2x(x^2+4)^{100}dx$$ Now you could simplify out the expression to the hundredth power and then take the integral of each individual part but I certainly don't want to do that. Instead, we can use u-substitution.

We do this by taking part of the expression inside the integral and setting it equal to a variable $u$. In this problem, we will have $u$ equal the expression inside the parentheses. $$u=x^2+4$$ The trick for u-substitutions is to look for things that are derivatives of one another. Notice that if you took the derivative of $x^2+4$ you'd get $2x$ which is what's outside the parentheses.

It derivative doesn't have to be exactly the other stuff, i.e. $2x$, but it's usually a clue for you to apply the u-substitution method to an integral.

Now, once you label what $u$ is, we want to take the derivative of $u$. Also, make sure you attach a $dx$ onto the end of it. $$\begin{aligned} &u=x^2+4\\ &du=2x\text{ }dx \end{aligned}$$

Once you have both $u$ and $du$, we want to substitute these into the integral. $$\begin{aligned} &\int 2x(\color{red}{x^2+4})^{100}dx\\\ &\int 2x(u)^{100}dx\\\ &\int \color{red}{2x}(u)^{100}\color{red}{dx}\\\ &\int u^{100}du \end{aligned}$$

We've successfully applied u-substitution and would you look at that, the integral is easier to solve. $$\int u^{100}du=\frac{1}{101}u^{101}+C$$

The final step is to substitute back in the expression that's equal to $u$. $$\boxed{\frac{1}{101}(x^2+4)^{101}+C}$$

Let's look at another example. $$\int(\cos^4\theta)\sin\theta\text{ }d\theta$$

This can be rewritten as: $$\int(\cos\theta)^4\sin\theta\text{ }d\theta$$

Usually, if an expression is being raised to a power, we want that expression to equal $u$. So we have: $$u=\cos\theta$$

Now, take the derivative of $u$. Don't forget to tack on the $d\theta$. $$\begin{aligned} &u=\cos\theta\\ &du=-\sin\theta\text{ }d\theta \end{aligned}$$

Notice this derivative is not the same as the remaining part in the integral, i.e. $-\sin\theta\ne\sin\theta$. This is okay. We can multiply both sides of this derivative by negative one in order to get the same expression in the integral. $$du=-\sin\theta\text{ }d\theta\Longrightarrow-du=\sin\theta\text{ }d\theta$$

We can now substitute these into the integral. $$\begin{aligned} &\int(\color{red}{\cos\theta})^4\sin\theta\text{ }d\theta\\\ &\int(u)^4\sin\theta\text{ }d\theta\\\ &\int(u)^4\color{red}{\sin\theta\text{ }d\theta}\\\ &\int(u)^4(-du)\\\ &-\int u^4du\\\ \end{aligned}$$

Finally, we can solve the interval. $$-\int u^4du=-\frac15u^5+C$$

Now just substitute back in the expression equal to $u$ and we have our answer. $$\boxed{-\frac15(\cos\theta)^5+C}$$

In the final example, Patrick looks at the integral: $$\int\frac{x}{\sqrt[4]{x+2}}dx$$

This can be rewritten as: $$\int\frac{x}{(x+2)^{1/4}}dx$$

Again, the expression being raised to the power will usually be the choice for $u$. $$u=x+2$$

Let's take the derivative of this. $$\begin{aligned} &u=x+2\\ &du=1\text{ }dx \end{aligned}$$

Notice though that we still have the $x$ in the numerator to deal with. We can actually use our equation with $u$ to find this. $$\begin{aligned} u&=x+2\\ u-2&=x \end{aligned}$$

Now, we can substitute these into the integral. $$\begin{aligned} &\int\frac{\color{red}{x}}{(x+2)^{1/4}}dx\\\ &\int\frac{u-2}{(x+2)^{1/4}}dx\\\ &\int\frac{u-2}{(\color{red}{x+2})^{1/4}}dx\\\ &\int\frac{u-2}{(u)^{1/4}}dx\\\ &\int\frac{u-2}{u^{1/4}}\color{red}{dx}\\\ &\int\frac{u-2}{u^{1/4}}du\\\ \end{aligned}$$

We can simplify this: $$\int\frac{u-2}{u^{1/4}}du=\int \left(u^{3/4}-2u^{-1/4}\right)du$$ Then we can split this into two integrals. $$\int \left(u^{3/4}-2u^{-1/4}\right)du\Rightarrow\int u^{3/4}du-\int 2u^{-1/4}du$$

Solve each integral separately then combine. $$\int u^{3/4}du=\frac47u^{7/4}+C\\\ -\int 2u^{-1/4}du=-\frac83u^{3/4}+C$$

And then combine: $$=\frac47u^{7/4}-\frac83u^{3/4}+C$$

Finally, we plug back in the $x$ expression for $u$. $$\boxed{\frac47(x+2)^{7/4}-\frac83(x+2)^{3/4}+C}$$

(Video) More Complicated Examples of U-Substitution

by PatrickJMT

In this video by patrickJMT he shows you how to solve some more complicated examples of integrals involving u-substitution.

The first integral he looks at is: $$\int e^{x+e^x}dx$$

Before we start, we can actually rewrite this expression. $$\int e^xe^{e^x}dx$$

Yes, we should set $u$ equal to $e^x$. $$u=e^x$$

We then also find the derivative. $$\begin{align} &u=e^x\\ &du=e^x\text{ }dx \end{align}$$

Now we can plug these into the integral. $$\begin{aligned} &\int \color{red}{e^x}e^{e^x}\color{red}{dx}\\\ &\int e^{e^x}du\\\ &\int e^{\color{red}{e^x}}du\\\ &\int e^{u}du\\\ \end{aligned}$$

Solve the integral. $$\int e^{u}du=e^u+C$$

Finally, plug back in the expression $u$ equals. This will be your answer! $$\boxed{e^{e^x}+C}$$

The next integral looked at is: $$\int\cot x\left(\ln\sin x\right) dx$$

This one might not be as obvious.

Yes, we should set $u$ to: $$u=\ln\sin x$$

If it's not apparent it becomes so when you take the derivative. $$\begin{aligned} &u=\ln\sin x\\ &du=\frac{1}{\sin x}\cdot\cos x=\cot x \end{aligned}$$

Next, plug these into the integral. $$\begin{aligned} &\int\color{red}{\cot x}\left(\ln\sin x\right)\color{red}{dx}\\\ &\int\ln\sin x\text{ }du\\\ &\int\color{red}{\ln\sin x}\text{ }du\\\ &\int u\text{ }du \end{aligned}$$

Solve the integral. $$\int u\text{ }du=\frac12u^2+C$$

Finally, we substitute back in the expression equal to $u$. $$\boxed{\frac12\left(\ln\sin x\right)^2+C}$$

Next up is the example: $$\int\frac{x}{1-x^2+\sqrt{1-x^2}}dx$$

Yes, $1-x^2$. $$u=1-x^2$$

Now take the derivative. $$\begin{aligned} &u=1-x^2\\ &du=-2x\text{ }dx \end{aligned}$$

We can rewrite this derivative in order for it to match the numerator of the integral. $$du=-2x\text{ }dx\Longrightarrow-\frac12du=x\text{ }dx$$

Plug these into the integral now. $$\begin{aligned} &\int\frac{\color{red}{x}}{1-x^2+\sqrt{1-x^2}}\color{red}{dx}\\\ &\int-\frac12\frac{1}{1-x^2+\sqrt{1-x^2}}du\\\ &-\frac12\int\frac{1}{\color{red}{1-x^2}+\sqrt{\color{red}{1-x^2}}}du\\\ &-\frac12\int\frac{1}{u+\sqrt{u}}du\\\ \end{aligned}$$

Hmm, what now? What if I told you we could use u-substitution all over again! Yeah, I'm groaning too. This time we label it $v$. $$v=\sqrt u$$ We can rewrite this as: $$v^2=u$$ Now, take the derivative. $$2v\text{ }dv=du$$

Plug these back into the derivative. $$\begin{aligned} &-\frac12\int\frac{1}{\color{red}{u}+\sqrt{u}}du\\\ &-\frac12\int\frac{1}{v^2+\sqrt{u}}du\\\ &-\frac12\int\frac{1}{v^2+\color{red}{\sqrt{u}}}du\\\ &-\frac12\int\frac{1}{v^2+v}du\\\ &-\frac12\int\frac{1}{v^2+v}\color{red}{du}\\\ &-\frac12\int\frac{1}{v^2+v}(2v)dv\\\ &-\int\frac{1}{v+1}dv\\\ \end{aligned}$$

This, we can integrate. $$-\int\frac{1}{v+1}dv=-\ln(v+1)+C$$

Now plug back in the $u$ expression. $$-\ln(\sqrt u+1)+C$$

Finally, plug back in the $x$ expression. $$\boxed{-\ln(\sqrt{1-x^2}+1)+C}$$

Lastly, he looks at: $$\int x\sqrt[3]{x+4}\text{ }dx$$

Yes, we want it to be the entire root not just what's inside. $$u=\sqrt[3]{x+4}$$ We can now rewrite this as: $$\begin{aligned} u&=\sqrt[3]{x+4}\\ u^3&=x+4\\ u^3-4&=x \end{aligned}$$

Take the derivative of this equation. $$3u^2\text{ }du=dx$$

Let's plug all these parts into the integral. $$\begin{aligned} \int \color{red}{x}\sqrt[3]{x+4}\text{ }dx\\\ \int (u^3-4)\sqrt[3]{x+4}\text{ }dx\\\ \int (u^3-4)\color{red}{\sqrt[3]{x+4}}\text{ }dx\\\ \int (u^3-4)(u)\text{ }dx\\\ \int (u^3-4)(u)\text{ }\color{red}{dx}\\\ \int (u^3-4)(u)(3u^2)\text{ }du\\\ \int (3u^6-12u^3)du\\\ \end{aligned}$$

We can now split this integral into two parts. $$\int (3u^6-12u^3)du\Rightarrow3\int u^6\text{ }du-12\int u^3\text{ }du$$

Solving each individually, we get: $$\begin{aligned} &=3\frac17u^7-12\frac14u^4+C\\\ &=3\frac17u^7-3u^4+C\\\ &=3\left(\frac17u^7-u^4\right)+C\\\ \end{aligned}$$

Finally, we can plug back in the expression equal to $u$. $$\boxed{3\left[\frac17\left(\sqrt[3]{x+4}\right)^7-\left(\sqrt[3]{x+4}\right)^4\right]+C}$$