Question

If u and v are two functions of x, then prove that:
\[\int{uv'dx}=u.v-\int{vu'dx}\]

Answer 1

Hint: The derivation of the integration of product of two differentiable functions is done using the product rule of differentiation. Let us assume that $ y=u.v $ then differentiate with respect to x on both the sides using product rule we get, $ \dfrac{dy}{dx}=\dfrac{d\left( u.v \right)}{dx}=u.\dfrac{dv}{dx}+v.\dfrac{du}{dx} $ . After differentiation, integrate on both the sides with respect to x and hence, you will get the required result.

Complete step-by-step answer:
It is given that we have to prove:
 $ \int{udv}=u.v-\int{vdu} $
It is also given that u and v are functions of x.
To prove the above integration, we can use the differentiation of products of two functions.
Let us take two functions u(x) and v(x) and let us assume that the product of these two functions is equal to y.
 $ y=u\left( x \right).v\left( x \right) $
Differentiating on both the sides we get,
 $ \dfrac{dy}{dx}=\dfrac{d\left( u\left( x \right).v\left( x \right) \right)}{dx}=u\left( x \right)\dfrac{dv\left( x \right)}{dx}+v\left( x \right)\dfrac{du\left( x \right)}{dx} $
In the above equation instead of writing u(x) and v(x) we can write only u and v. The meaning is the same but we have shortened the notation so that the calculation will be easy. Rewriting the above equation in terms of u and v we get,
 $ \dfrac{d\left( u.v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx} $
Now, integrating on both the sides we get,
 $ \int{\dfrac{d\left( u.v \right)}{dx}dx=\int{u\dfrac{dv}{dx}}dx+\int{v\dfrac{du}{dx}}}dx $
We know that differentiation followed by integration will give the original number on which differentiation and integration has been applied so we can write the above equation as:
 $ u.v=\int{u\dfrac{dv}{dx}dx+\int{v\dfrac{du}{dx}dx}} $
In the above equation, we can write $ \dfrac{dv}{dx}=v' $ and $ \dfrac{du}{dx}=u' $ .
 $ u.v=\int{uv'dx+\int{vu'dx}} $
Rearranging the above equation we get,
\[\int{uv'dx}=u.v-\int{vu'dx}\]
Hence, we have proved what has been asked in the problem.

Note: The interesting point in this problem is that the integration that we have derived is the derivation of integration by parts. The other form of integration by parts is shown below in which when two functions u and v are given in the product form and we have to find its integration.
\[\int{u.vdx=u\int{vdx-\int{u'\left( \int{vdx} \right)dx}}}\]
In the above formula, the point to be taken care of is the integral written after the negative sign on the right hand side of the above equation. In the following integral:
\[\int{u'\left( \int{vdx} \right)dx}\]
To evaluate this integral, first find the integration of $ \int{vdx} $ then put the result in the above expression and then integrate the above expression. Generally students get confused with this part of the integration, so they must be very clear.