Subject: Calculus

Integration By Parts

Doing integration is straightforward especially when the function to be integrated can be expressed as a single integrable function. But when the integration looks like below

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

errors are commonly committed especially for those who aren't knowledgeable about integration by parts. Integration by parts is a technique in integration that solves the integral of the types of functions depicted above. The integral above is not as easy as it seems without doing integration by parts and thus possesses difficulty among first-time calculus students or learners.

Common integration errors...

Most of the common errors made in evaluating the above integral is trying to solve it like this;

\int f(x)g(x)dx=\int f(x)dx \int g(x)dx

which is completely false, of course. When you make g=2 and say f=1, then the integral would become 2x=2x^2 which is completely not true!

How do we correct such errors?

One well-known solution to such integral is integration by parts. Integration by parts works by transforming the integral of products of functions into other (ideally simpler) integrals. Consider the following:

D[f(x)g(x)] = f(x)\cdot g'(x)+g(x)\cdot f'(x)

We know that the above equation is the so-called product rule of the derivatives of a function. Now, if we integrate both sides of the function above,

\int D[f(x)g(x)] = \int f(x)\cdot g'(x)dx+\int g(x)\cdot f'(x)dx

This gives us

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

Rearranging, we get

\int f(x)\cdot g'(x)dx =f(x)g(x) - \int g(x)\cdot f'(x)dx

Letting u=f(x) such that du=f'(x)dx and v=g(x) such that dv=g'(x)dx, the equation above would then take the form of

\int u\cdot dv = uv - \int v \cdot du

This would be our formula for solving the integral of products of functions. And of course, this is what we call the integration by parts method. The nice thing about this equation is when one of the two integrals in f'(x)dx and g'(x)dx can be integrated easily, then we can use it to find other the remaining integrals by transforming it into an integral like the one above.

Let us learn integration by parts step-by-step

Doing integration by parts is not a "mechanical process" where a certain set of rules or methods or way must be followed. Rather, integration by parts requires careful analysis, critical thinking and proper choice of the functions to be assigned by v or du which leads me to the first step;

First Step

Basically. integration by parts is performed when you have an integral such as \int f(x)g(x)dx. Then, it can be transformed into a form \int u\cdot dv by carefully selecting which function is assigned with u and dv. Make sure that the assigned differentials correspond also to a differential. For example, you can assign f(x)=u and g(x)dx=dv or g(x)=u and f(x)dx=dv such that the integral \int f(x)g(x)dx is now of the form \int u\cdot dv .

Second Step

Make sure that when you assign functions, the term g(x)dx or f(x)dx is easily integrable. Then if it is, integrate f(x)dx=dv or g(x)dx=dv to attain a value of v

Third Step

After that, differentiate whichever function assigned with u. So you will have either du=f'(x) or du=g'(x).

Fourth Step

From the second step, you attained the value of v and du in the third step. Then, integrate

\int v \cdot du

So, after that, insert them into our formula presented above, which I repeat, is

\int u\cdot dv = uv - \int v \cdot du.

And you get the value of the integral \int u\cdot dv which is just the same to the integral \int f(x)g(x)dx.

Illustration

No other perfect illustration for methods like these is very appealing than providing a concrete example of doing integration by parts. Suppose you are to integrate x\cos{x}dx. This can never integrate directly so we use integration by parts.

First step

The is of the form x\cos{x}dx and let's choose which one is u and which one is dv. Let us remember that we need to carefully choose these values so that after third step, we can evaluate \int v \cdot du. If we choose dv=x and u=\cos{x}, then the integral \int v \cdot du becomes \int -x^2 \cdot \sin{x} dx which is far more difficult than our original integral which is x\cos{x}dx. So, let's choose dv=\cos{x} and u=x.

Second step

Since we choose dv=\cos{x}, let's integrate it,

dv=\cos{x}

\int dv=\int \cos{x}

v=\sin{x}

Third step

We'll differentiate u. That is

u=x
du=dx

Fourth step

Let's evaluate the integral \int v \cdot du. Since we have du=dx and v=\sin{x}, then

\int v du
\int \sin{x} dx
\cos{x}

Then, inserting them into

\int u\cdot dv = uv - \int v \cdot du,

we get

\int x\cos{x}dx= x\sin{x}- \int \sin{x} dx
\int x\cos{x}dx= x\sin{x}- \cos{x} dx

So using integration by parts, we have successfully solved our integral.

Example #1

Calculate: \int \ln{x} dx

ANSWER: \int \ln{x} dx = x\ln{x}-x + C

Example #2

Integrate using integration by parts: \int x\ln{x} dx

ANSWER: \int x\ln{x} dx = \frac{1}{2}x^2 \ln{x}-\frac{1}{4}x^2 + C

Example #3

Solve: \int e^x \sin{x}dx

\int e^x \sin{x}dx = \frac{1}{2}e^x (\sin{x}-\cos{x})+C

Example #4

Calculate: \int e^x\cos{x}dx

\int e^x\cos{x}dx = \frac{1}{2}e^x(\sin{x}+\cos{x})+C

Example #5

Calculate: \int x\sqrt{x+1}dx

\int x\sqrt{x+1}dx=\frac{2}{3}x(x+1)^{3/2}-\frac{4}{15}(x+1)^{5/2}+C

NEXT TOPIC: Inverse chain rule method