#u-substitution

[lily-of-the-flame]

Can someone explain U-Substitution to me? I can never seem to get it, which is annoying because I'm generally good at math.

[balajirao1980-blog]

#30. INTEGRATION BY SUBSTITUTION | INTEGRATION | U-SUBSTITUTION

[balajirao1980-blog]

#22 INTEGRATION BY U-SUBSTITUTION | U-SUBSTITUTION | INTEGRATION BY SUBS...

[nick1100]

How to solve integrals with u-sub and why it works

Finding a primitive, or antiderivative, of a function is at times hard. But there are some important and well-known techniques for doing that, and one of them is the u-substitution technique. You could say that it's based on the chain rule. If you want to refresh your memory on that, I've blogged about it in another post.

Perhaps you've seen before that, if you’ve have an integral of the following kind

$$\int f(g(x)) g'(x) \, dx $$

and we let $ u = g(x) $ it can be (hopefully) be rewritten to a more managable integral

$$ \int f(u) \, du. $$

But why it works is very interesting. How fun is it to use something without knowing why it works the way it does? ;) I said in the beginning that you could say this technique comes from the chain rule. To get an idea of why it works (a proof could be made using the fact that chain rule is true), recall that, the chain rule tells us what happens if we want to figure out the derivative of a composition of two or more functions:

$$ \frac{df}{dx} f(g(x)) = f'(g(x))\cdot g'(x)) $$

And we also know that, in order to get back to $ f(g(x)) $ we just need to integrate $ f'(g(x))\cdot g'(x)) $ with respect to $ x $, e.g.

$$ \int f'(g(x))\cdot g'(x)) \,dx= f(g(x)) + C $$

Usually we let $ F $ be the antiderivative of $ f $, and also note that it doesn't really matter what we call our functions, so for pedagogical reasons I will just say that $ g(x) = u(x) $:

$$ \int f(u(x))\cdot g'(x))\, dx = F(u(x)) + C $$

Now compare this with the following formula

$$ \int f(u) \, du = F(u) + C $$

If we say that $ u = u(x) $ and $ du = g'(x) \, dx $ we end up with the exact same thing! So you can indeed say that, the u-substitution technique works because of the fact that the chain rule works.

Some teachers probably tell you to imagine that $ \frac{dy}{dx} $ is a fraction and not an operator. If you want to do that, it's fine - as long as you feel confident. But now you probably have an idea on why you can "pretend" that it is an fraction and treat it that way.

Example

Compute the indefinite integral

$$ \int \, 2x \ sin\,x^2 \, dx $$

It's not hard to see that if we let $ u = x^2 $ then it's derivative $ u' = 2x $. If we now treat the $ \frac{dy}{du} $ operator as a fraction (remember that we said it was "legal" in this case) we would get

$$ \frac{du}{dx} = 2x \Leftrightarrow dx = \frac{du}{2x} $$

So now we also know what $ dx $ is, we get

$$ \int \, 2x \ sin\,(u) \, \frac{du}{2x} $$

Notice that, $ 2x $ cancels out, thus

$$ \int \, 2x \ sin\,u \, \frac{du}{2x} = \int \, sin\,u \, du $$

Now it's easy to see the antiderivative, namely

$$ \int \, sin\,u \, du = -cos \, u + C $$

Remember we had the variable substitution $ u = x^2 $ so we substitute it back, thus our answer is

$$ -cos \, x^2 + C $$

More examples can be found in this post

[nick1100]

Integrating slightly more complex functions examples

It’s clear that to solve integrals all we need to do is to find the antiderivative. However, sometimes the antiderivative is not so obvious and sometimes it’s not even possible to tell, hence the need for numerical methods. In this post  we’ll use some techniques like u-substitution and partial integration and we can see how to approximate integrals in another one. Let’s start with an easy example. 

Example 1

Solve the indefinite integral

$$ \int 2x \, e^{x^2} \, dx. $$

Clearly the antiderivative is not so obvious right away. What we want to do is to rewrite that integral to be more manageable. Let me show you how we can solve this first, and then we’ll discuss why. 

If we let $ u $ be $ x^2 $ we can see that it’s derivative, $ 2x $, is the same as the factor $ 2 x $ in the integral. This means that if we take the derivative of $ x^2 $ and solve for $ dx $ we’ll able to get rid of that, and we’ll be left with a very easy integral.

Again, let $ u $ be $ x^2 $ in 

$$ \int 2x \, e^{u} \, dx. $$

The derivative of $ u $, $ u’ $, is given by

$$ \frac{du}{dx} = 2x \Leftrightarrow dx = \frac{du}{2x}.$$

Since we have solved for $ dx $ we get

$$ \int 2x \, e^u \, dx = \int 2x \, e^u \frac{du}{2x} = \int e^u \, du = e^u + C $$

Since $ u = x^2 $ we get that $$  e^{x^2} + C $$

thus

$$  \int 2x \, e^{x^2} \, dx =  e^{x^2} + C $$

where $ C $ is an arbitrary constant.

Example 2

Solve the indefinite integral

$$ \int \frac{lnx}{x} \, dx. $$

If we apply same logic here as before, if we let $ u = ln x $ we get that $$ \frac{du}{dx} = \frac{1}{x} \Leftrightarrow dx = x \, du $$

which lets us rewrite the original integral to a more manageable one and get rid of that irritating $ x $ in the denominator. In other words, $ u = ln x $ seems like a good substitution. We get that 

$$ \int \frac{lnx}{x} dx = \int \frac{u}{x} x \, du = \int u \, du = \frac{1}{2} u^2 + C    $$

substitute $ u $ back to $ ln x $ and we get that

$$ \int \frac{lnx}{x} dx = \frac{1}{2} (lnx)^2 + C $$

where $ C $ is an arbitrary constant.

Example 3 

Solve the indefinite integral

$$ \int cos \, 5x \, dx. $$

Let $ u $ be $ 5x $ in 

$$  \int cos \, u \, dx. $$

Since $ \frac{du}{dx} = 5 $ then we have that $ dx = \frac{du}{5} $, thus

$$ \int cos \, 5x \, dx = \int cos \, u \, \frac{1}{5} du = \frac{1}{5} \int cos \, u \, du = \frac{1}{5} sin \, u + C $$

Substitute $ u $ back and we get our answer:

$$ \int cos \, 5x \, dx =  \frac{1}{5} sin \, 5x \, + C.  $$ 

Note:

We could actually have solved the last one without having to do the substitution. If we think about it, the antiderivative of $ cos \, x $ is simply $ sin x + C $ since we get back again to  $ cos \, x $ if we take the derivative of  $ sin x $, in same way as the antiderivative of $ sin \, x $ is $- cos \, x + C $. Similarly if we have $ cos 5x $ the antiderivative must be  $ \frac{1}{5} sin \, 5x  \, + C $ because of the chain rule. Generally, we could say that

$$ \int cos \, ax \, dx = \frac{1}{a} sin \, ax + C $$

and

$$ \int sin \, ax \, dx = -\frac{1}{a} cos \, ax + C $$

for some constant $ a \neq 0 $ and $ C $ is an arbitrary constant.

Example 4

 Compute the definite integral

$$ \int_{0}^{2} \sqrt{3x + 1} dx. $$

It’s obvious that if we take the derivative of $ 3x + 1 $ we’ll get a constant which we simply can multiply the integral with, so letting $ u $ be $ 3x + 1 $ is a good choice.

Let $ u = 3x + 1 $ which gives us that $ \frac{du}{dx} = 3 \Leftrightarrow dx = \frac{du}{3}. $

We get

$$ \int_{0}^{2} \sqrt{3x + 1} dx = \int_{x = 0}^{x = 2} \sqrt{u} \frac{du}{3}   = \frac{1}{3}  \int_{x = 0}^{x = 2} \sqrt{u} \,du = $$

$$ =   \frac{1}{3} \left[\frac{u^{3/2}}{3/2}\right]_{x=0}^{x=2} = \frac{1}{3} \left[2\frac{u^{1/2 + 1}}{3}\right]_{x=0}^{x=2}   = \frac{1}{3}     \left[2\frac{\sqrt{3x+1}(3x+1)}{3}\right]_{0}^{2} = $$

$$ = \frac{1}{3}(\frac{14\sqrt{7}}{3} - \frac{2}{3}) = \frac{14\sqrt{7} - 2}{9}.    $$ 

Note:

If you change the integration limits you wouldn’t have to substitute back like I did - you could just have worked with $ u $ instead. You calculate the limits by “plugging” your x values into your “substitution function”. For instance, mine would have been $ u = 3x + 1 = 2 \cdot 3 + 1 = 7 $ and $ u = 3x + 1 = 3 \cdot 0 + 1 = 1 $.

Example 5

Solve the indefinite integral

$$ \int 3x \, e^{-x} \, dx $$

In this one there’s not really a sign of that we can use u-substitution. However, the multiplication reminds of the technique integrating by parts, which is inspired by the product rule. The formula tells that

$$ \int f(x) \cdot g(x) \, dx = F(x) \cdot g(x) - \int F(x) \cdot g’(x) \, dx $$

If $ f(x) = e^{-x} $ and $ g(x) = 3x $ then $ F(x) = -e^{-x} $ and $ g’(x) = 3. $

Why did I choose those functions? Well, I’m reasoning that if I put $ g(x) = 3x $ I will get rid of the x factor and be left with a constant, which simplifies a lot.

Now we have that

$$ \int 2x \, e^{-x} \, dx = -e^{-x} \cdot 3x - \int -e^{-x} \cdot 3 \, dx = -e^{x} 3x - 3e^{-x} = -3e^{-x}(x + 1). $$

Thus

$$ \int 2x \, e^{-x} \, dx = -3e^{-x}(x + 1).  $$

Example 6

 Solve the indefinite integral 

$$ \int x^2 sin \, x \, dx. $$

If we let $ f(x) = sin \, x $ and $ g(x) = x^2 $ then we get $ F(x) = -cos \, x $ and $ g’(x) = 2x $.

So we get 

$$ \int x^2 sin \, x \, dx = -cos \, x \cdot x^2 - \int -cos \, x \cdot 2x \, dx.  $$

This helped only a bit, since we still can’t take the antiderivative. However, if we integrate again by parts and this time put $ f(x) = -cos \, x $ and $ g(x) = 2x $ we get

$$  -cos \, x \cdot x^2 - \int -cos \, x \cdot 2x \, dx = $$ $$ = -cos \, x \cdot x^2 - (-sin \, x \cdot 2x - \int -sin \, x \cdot 2 \, dx) =  $$ $$ =   -cos \, x \cdot x^2 - ( -sin \, x \cdot 2x - 2cos\, x) = $$ $$ = -cosx \cdot x^2 + sin\, x \cdot 2x + 2cos \, x. $$

Thus, $$  \int x^2 sin \, x \, dx = 2x \, sin\, x + 2cos \, x - x^2 \, cos \, x \,  $$

[notcharting2day]

AP:)))Cal:)))cu:))))lus:)))) is :)) muc:))h fun

[nathanielbuildsatesseract]

It's been so long since I took a derivative. How the hell does u-substitution work again?

[oniongalaxy]

U-substitution is just the chain rule in reverse.

The integral of (sinx)^3 * cosx = (1/4)sinx^4 using the chain rule in reverse.

Because the chain rule in reverse is:

integral sign (g'(x) * (f(g(x))) ) = f (g(x))

This is why u - substitution makes sense. Because essentially u=sinx and du=cosx,

giving you:

integral sign u^3 * du = 1/4 u^4 or = ( 1/4)sinx^4

My teacher never told me this, so I thought I'd share it with you all.