Integration is a very important computation of calculus mathematics. Many rules and formulas are used to get integration of some functions. A special rule, which is integration by parts, is available for integrating the products of two functions. This topic will derive and illustrate this rule which is Integration by parts formula. Also, some examples will help the students to get their concept. Let us start!

**Integration by Parts Formula**

**What is integration by parts method?**

This method is very useful in order to master the technique of integrations. Many times we have to integrate the product of two functions. Functions often arise as the products of other functions, and so we have to integrate these products. For example, we may be asked to determine

\(\int x\); \(\cos x\); dx

Here, the integrand is the product of the two functions x and cos x. A rule exists for integrating the products of functions which is required for getting the solution.

**Derivation of the formula for integration by parts:**

This rule states that:

**\( \(\int {u\frac{{dv}}{{dx}}}\) dx = uv – \int {\frac{{du}}{{dx}}} vdx \)**

**Derivation: **If y = uv

As we know that,

\(\frac{dy}{dx}\)= \(\frac{d}{dx}\) uv = u \(\frac{d}{dx}\) v + v \(\frac{d}{dx}\) u

Rearranging it,

\( u\frac{d}{dx}v= \frac{d}{dx}uv – v\frac{d}{dx}u \)

Now integrating both sides,

\(\int u\frac{dv}{dx} dx= \int \frac{d(uv)}{dx} dx – \int v\frac{du}{dx} dx \)

The first term on the right hand side simplifies since we are simply integrating what has been differentiated.

**\( \int u\frac{dv}{dx} dx= uv – \int v\frac{du}{dx} dx \)**

This formula is known as integration by parts. This formula is very useful for solving complex integration problems. In some questions, we will see that it is sometimes necessary to apply the formula for integration by parts more than once.

This gives us the method for integration, called INTEGRATION BY PARTS. This method allows us to integrate many products of functions of x. We take one factor in the product as u (this also appears on the right-hand side, along with \(\frac{du} {dx} ) \).

The other factor is taken to be \(\frac{ dv}{dx}\) (on the right-hand-side the only v appears)

**Solved Examples**

**Q:** Solve the integration given:

\(\int x\; cosx \;dx \)

**Solution: **

Here, we are trying to integrate the product of two functions x and cos x. To use the integration

by parts method we let one of the terms be

\(\frac{dv}{dx}\) and the other be u.

See from the formula that whichever term we let equal u we need to differentiate it in order to find \(\frac{du}{dx}\)

So in this case, if we assume u as x, so when we differentiate it we will find

\(\frac{du}{dx} = = 1, \)

i.e. simply a constant. Notice that the formula will replace one integral, the one on the left, by another, and one on the right. Careful selection of u will produce an integral which is less complicated than the original.

Thus

\(u= x and \frac{dv}{dx} = cosx \)

With this choice, by differentiating we obtain

\(\frac{du}{dx} = 1\)

Also from \(\frac{dv}{dx} = cos x, by integrating we find\)

\(v= \int cosx \; dx = sinx \)

Then use the formula,

\(\int {u\frac{{dv}}{{dx}}} dx = uv – \int {\frac{{du}}{{dx}}} vdx \)

\( \int x\;cosx\;dx = x\;sinx – \int (sinx).1\;dx \)

\(= x\;sinx+cosx+c \) , which is the solution

Where c is constant of integration.

I get a different answer for first example.

I got Q1 as 20.5

median 23 and

Q3 26

Hi

Same

yes