U-Substitution

$U-Substitution Example 1$

$U-Substitution Example 2$

$U-Substitution Example 3$

$U-Substitution Example 4$

$U-Substitution Example 5$

$U-Substitution Example 6$

$U-Substitution Example 7$

PJ Integration by U-Substitution: Antiderivatives

PJ Integration by U-Substitution – Indefinite Integral, Another 2 Examples

PJ Integration by U-substitution, More Complicated Examples

U-Substitution is a method to find anti-derivatives, in other words, a method to do integrations.
Up to now, we already know how to do the following integrals by using anti-derivatives, such as
$\int x^n dx=\frac{1}{n+1}x^{n+1}+c$ , when $n\neq-1$ ,
since $\frac{d x^{n+1}}{dx}=(n+1)x^n$ ;
$\int \frac{1}{x} dx=\ln(x)+c$ ,
since $\frac{d \ln(x)}{dx}=\frac{1}{x}$ ;
$\int \sin(x)dx=-\cos(x)+c$ ,
since $\frac{d \cos(x)}{dx}=-\sin(x)$ ;
$\int \cos(x)dx=\sin(x)+c$ ,
since $\frac{d \sin(x)}{dx}=\cos(x)$ ;
$\int e^x dx =e^x+c$ ,
since $\frac{d e^x}{dx}=e^x$ , and
$\int a^x=\frac{a^x}{\ln a}a^x+c$
since $\frac{d a^x}{d x}=\ln a\cdot a^x+c$
where in all the formula above, we denote the integration constant by $c$ .
It is to say that we have already mastered how to do integration when we can recognize the integrand as a derivative for some function.
Further we are also equipped with the knowledge that the operation of integration is linear, that is,
$\int (c_1 g(x) +c_2h(x))dx=c_1\int g(x)dx+c_2\int h(x)dx$ , where $c_1, c_2$ are constants and $g(x), h(x)$ are integrable functions.
Now, by using the method of u-substitution, we will be able to integrate functions that we have a harder time to immediate recognize as derivatives. The idea of u-substitution is to use algebraic manipulations to simplify a form of a function such that its antiderivative will be easily identified.
Let us start out with trying to do the following integral,
$\int 4e^{4x+3}dx$ .
What is our strategy here? Let us review the chain rule: if $y=f(u), u=g(x)$ , then $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$ .
Using this, and $\frac{d e^x}{dx}=e^x$ , we can perform the following derivative operation,
$\frac{d e^{4x+3}}{dx}=e^{4x+3}\frac{d (4x+3)}{dx}=4e^{4x+3}$ ,
which we realize the result is exactly the integrand of the integral we need to find. So we immediately have
$\int 4e^{4x+3}dx=e^{4x+3}+c$ .
So how will we do if we use the method of u-substitution? We apply the chain rule backwards! So here is our substitution rule in general:
$y(x) =\int y'(u(x))u'(x) dx+c$ , where we use $‘$ to denote derivative respect to the argument.
Let us illustrate how to do that on this particular example. The integral we want to do is
$\int 4e^{4x+3}dx$ ,
we want to simplify the integrand, let
$u=4x+3$ , then we have
$\frac{du}{dx}=4$ . Treat this as an algebraic equation, then we have $du=4dx$ , now we are ready to do the substitution,
$\int 4e^{4x+3}dx=\int 4e^udx=\int e^udu=e^u+c=e^{4x+3}+c$
In this simple example, we illustrate how to use the method of u-subsitution, in other words, using the chain rule backwards.
U-subsitution is very useful in the cases when the integrand is not a simple function of x, such as not just $x^a$ , but with a more complicated base; not just $e^x$ , but with a more complicated exponent; not just $\sin(x)$ or $\cos(x)$ , but with more complicated argument for the trig functions.
The best way to learn a method to integrate, is to use the method to do a lot of integration! Now let us look at some more examples.
1) $\int (x+2)e^{x^2+4x+3}dx$
For this problem, we immediate realize the exponent of the exponential function is very complicated, and we can try to substitute that to be $u$ and see what we get.
Solution: Let $u=x^2+4x+3$ , then we have $\frac{du}{dx}=2x+4$ , which is $du=(2x+4)dx=2(x+2)dx$ , now we substitute into the original integrand and obtain,
$\int(x+2)e^udx=\int e^u\cdot \frac{du}{2}=\frac{1}{2}\cdot\int e^udu=\frac{1}{2}e^u+c=\frac{1}{2}e^{x^2+4x+3}+c$ .
In this case, we realize that the derivative $du$ may not be able to absorb the entire integrand, and we need to use the linear property of the integration to obtain the right answer.
2) $\int (2x+3)(x^2+3x+4)^5dx$
The integrand for this integral is a polynomial function of x, so if we expand the expression, we will be able to integrate it, except that seems to be very complicated. Instead, we can also use u-substitution for this problem. By the spirit of u-substitution, we spot the inside function $x^2+3x+4$ could be substituted into $u$ which will make the function look much easier.
Solution: Let $u=x^2+3x+4$ , so we have $du=(2x+3)dx$ , hen we substitute to the original integral, and have
$\int (2x+3)u^5dx=\int u^5du=\frac{1}{6}u^6+c=\frac{1}{6}(x^2+3x+4)^6+c$
3) $\int \frac{2}{x}\cos(\ln x) dx$
This integrand has a $\ln x$ as the argument for the $\cos$ function, which looks complicated, so we immediately choose $\ln x$ as our u, for the purpose of making the integrand simpler.
Solution: Let $u=\ln x$ , so we have $du=\frac{1}{x}dx$ , then we have
$\int \frac{2}{x}\cos(u)dx=\int 2\cos(u)du=2(-\sin(u))+c=-2\sin(\ln x)+c$
4) $\int \frac{5x+15}{x^2+6x-9}dx$
Solution: Let $u=x^2+6x-9$ , and then we have $du=(2x+6)dx=2(x+3)dx$
Then, we have
$\int \frac{1}{u}5(x+3)dx=\frac{5}{2}\int \frac{1}{u}du=\frac{5}{2}\ln u+c=\frac{5}{2}\ln (x^2+6x-9)+c$
Now let us move on to some slightly more complicated integrals to do. For example, some of the integrands seem to have two of these inside functions, which seem to require two u-substitutions, like this one,
$\int \frac{\sin (4x)}{e^{2\cos(4x)}}dx$ .
First, we recognize the argument of the trig function is not simply $x$ , but $4x$ and then we realize the exponent of the exponential function is a trig function instead of $x$ as well. So there are two possibilities:
1) Doing two substitutions, letting $u=4x$ and $v=\cos(u)$
or
2) Doing only one substitution, letting $u=\cos(4x)$
Since the second way will only require one substitution instead of two, we choose to use only one substitution: we ignore the inside function $4x$ that is buried in the trig function and directly use
$u=2\cos(4x)$ , then we have $du=2\cdot 4\sin(4x)dx=8\sin(4x)dx$ , then the original integral becomes
$\int \frac{8\sin(4x)}{8}\cdot \frac{1}{e^u}dx=\int \frac{1}{8}e^{-u}du=-\frac{1}{8}e^{-u}+c=\frac{1}{8}e^{-2\cos(4x)}+c=\frac{1}{8e^{2\cos(4x)}}+c$
Let us summarize the steps of u-substitution:
Step one: We want to identify the function that we want to be substituted by $u$ , which is probably one of the following: a complicated base, or a complicated exponent, or a complicated argument of $\sin$ and $\cos$ .
Step two: We calculate the derivative of the function $u$ and obtain $du$ in terms of $late x$ and $dx$ , and keep in mind that we may also need $2du$ or $\frac{1}{2}du$ etc.
Step three: substitute $u$ and $du$ back in the original expression and check that all terms with $x$ can be eliminated! This step is very important, after u-substitution, there should be no term that contains $x$ .
Step four: integrate in terms of $u$ , which should be a simple integral to do, and substitute back to $x$ .

U-Substitution

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