What if the Same Integral Appears Again After Using Product Rule of Differentiation

When we first larn how to integrate, the examples we run into involve simple polynomials, or single functions like these:

$\int{(x^2+4)}\,dx=\frac{x^3}{3}+4x+C$

$\int{\sin{x}}\,dx=-\cos{x}+C$

Integrals of products

What if we need to detect the integral of a product of 2 functions, like the following example?

Instance one

$\int{{x}\,\sin{x}}\,dx$

This is where we need the important and useful technique in calculus known as integration by parts. (You lot tin can see a full caption starting from bones principles and with more examples here: Integration by parts).

To observe this integral, we cull "u" such that its derivative is simpler than u. In this case, nosotros volition choose u = x and proceed as follows:

u = x	dv = sin ten dx
du = dx	v = −cos ten

Nosotros apply the integration by parts formula and observe the integral:

$\int{u\,dv}=uv-\int{v\,du}$

$\int{x}\,\sin{x}\,dx=(x)(-\cos{x})-\int-\cos{x}\,dx$

Tidying this up gives:

$\int{x}\,\sin{x}\,dx=-x\,\cos{x}+\int\cos{x}\,dx$

Now, that final integral is like shooting fish in a barrel and we tin write our terminal answer:

$\int{x}\,\sin{x}\,dx=-x\,\cos{x}+\sin{x}+C$

Annotation 1: The constant of integration (C) appears later on nosotros do the final integration.

Note 2: Choosing u and dv can crusade some stress, but if you follow the LIATE rule, it is easier. For u, choose whatever comes highest in the folloentrwing listing, and cull dv as the lowest in this list.

L - logarithm functions
I - Inverse trigonometric functions
A - Algebraic functions (simple polynomial terms)
T - trigonometric functions
E - Exponential functions

Integration by parts - twice

Now, let's see a case that is double-barreled. That is, we don't get the answer with one round of integration by parts, rather nosotros need to perform integration past parts two times.

Example two:

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

In this example we choose u = x ² , since this will reduce to a simpler expression on differentiation (and it is higher on the LIATE listing), where e^x will non.

u = x ²	dv = e^x dx
du = 2x dx	v = e^ten

Now for integration by parts:

$\int{u\,dv}=uv-\int{v\,du}$

$\int{x^2e^x}dx={x^2}e^x-\int{e^x}(2x)\,dx$

Nosotros re-accommodate this to give the following, which I call equation [one]:

$\int{x^2e^x}dx={x^2}e^x-2\int{x}\,{e^x}\,dx\;\;\;[1]$

This time we tin't immediately practice that last integral, so nosotros need to perform integration by parts once again. Choosing "u" so that its derivative is simpler than u again, we accept:

u = x	dv = e ^x dx
du = dx	v = due east ^x

Note that the u and 5 hither accept different values from the u and v at the beginning of Example 2. This tin can be a trap if you lot don't write things carefully!

Now nosotros proceed using integration by parts on $\int{{x}\,e^x}dx$ :

$\int{u\,dv}=uv-\int{v\,du}$

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

That last integral is simple, and we go the following, which I call equation [2]:

$\int{{x}\,e^x}dx={x}\,e^x-e^x+C\;\;\;[2]$

Only we haven't finished the question - we must retrieve we are finding this integral:

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

This was our answer to the start integration by parts:

$\int{x^2e^x}dx={x^2}e^x-2\int{x}\,{e^x}\,dx\;\;\;[1]$

Substituting answer [2] this into equation gives united states of america:

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

Tidying this upward, we obtain the final answer:

$\int{x^2e^x}dx=e^x(x^2-2x+2)+C$

Notice the identify where the abiding "+C" appears in our reply - it's after thel integration has been performed. (Some students become hung up on this pace, or add the "+C" before information technology is appropriate, and some forget to add information technology at all!) I take used a subscript 1 on the first constant since information technology is not the same value equally the final C.

Integration by parts twice - with solving

Nosotros also come across integration by parts where nosotros really take to solve for the integral nosotros are finding. Here's an example.

Example iii:

$\int e^x\sin{x}\,dx$

In this instance, information technology is non so clear what we should cull for "u", since differentiating e¹⁰ does not give us a simpler expression, and neither does differentiating sin x . Nosotros choose the "simplest" possiblity, equally follows (even though eastward^x is below trigonometric functions in the LIATE table):

u = e¹⁰	dv = sin ten dx
du = eastward^x dx	5 = −cos x

Apply the integration past parts formula:

$\int{u\,dv}=uv-\int{v\,du}$

We obtain the following, which I'll call equation [3]:

$\int{e^x\sin{x}\,}dx=-e^x\cos{x}+\int{e^x\cos{x}\,}dx$

Now, for that terminal integral:

$\int{e^x\cos{x}\,}dx$

Once more, we need to decide which role to use for u, and settle on the one which gives simplest derivative:

u = e^x	dv = cos x dx
du = e^x dx	5 = sin x

Applying integration by parts for the second time:

$\int{u\,dv}=uv-\int{v\,du}$

We obtain equation [iv]:

$\int{e^x\cos{x}\,}dx=e^x\sin{x}-\int{e^x\sin{x}\,}dx$

Await a minute - nosotros have a final integral that is the same every bit what we started with! If nosotros kept going, we would go around in circles and never finish.

So we demand to perform the following "play tricks". Nosotros substitute our reply for the second integration past parts (equation [4]) into our starting time integration past parts respond (equation [3].

$\int e^x\sin{x}\,dx$

$=-e^x\cos{x}+\int e^x\cos{x}\, dx$

$=-e^x\cos{x}+\left[e^x\sin{x}-\int e^x\sin{x}\,dx\right]$

Removing the brackets:

$\int e^x\sin{x}\,dx=-e^x\cos{x}+e^x\sin{x}-\int e^x\sin{x}\,{dx}\ \ [2]$

Now, this equation is in the following course:

p = −q + r − p

To solve this for p, we simply add p to both sides:

2p = −q + r

Then divide both sides by two:

p = (−q + r)/2

So we will do the same to our integral equation, number [5].

I add $\int{e^x\cos{x}\,}dx$ to both sides:

$2\int e^x\sin{x}\,dx=-e^x\cos{x}+e^x\sin{x}+K$

Dividing both sides past 2 gives:

$\int e^x\sin{x}\,dx=\frac{e^x(\sin{x}-\cos{x})}{2}+C$

So we have solved equation [5] for $\int e^x\sin{x}\,dx$ , giving united states of america the desired effect.

(Note I used a "+1000" for the first constant that appeared. My concluding "C" has value Grand/ii, but normally nosotros only need to be concerned with the concluding constant.)

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Source: https://www.intmath.com/blog/mathematics/integration-by-parts-twice-5396

What if the Same Integral Appears Again After Using Product Rule of Differentiation

Integrals of products

Integration by parts - twice

Integration by parts twice - with solving

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