Question

What is the meaning of partial integration?

Answer 1

The name itself tells us the answer for the question. And also recall the concept of differentiating a product of functions by differentiating them partially (meaning differentiating them by parts). So we are going to implement the same concept here.
$\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}(v) + v\dfrac{d}{{dx}}(u)$ , where $u$ and $v$ are two different functions in $x$ .

Complete step-by-step solution:
In calculus, partial integration is a method of integrating a function which is a product of two different functions.
Suppose that, $u$ and $v$ are two different functions in $x$ , then,
$\int {uvdx = u\int {vdx} - \int {\left( {u'\int {vdx} } \right)dx} }$ .
Generally, this formula is written as,
$\int {uv'dx = uv - \int {\left( {u'v} \right)dx} }$ .-----( $1$ )
If we consider $u = u(x), \Rightarrow du = u'dx$ .
And similarly, $v = v(x), \Rightarrow dv = v'dx$
So equation $1$ changes as,
$\int {udv} = uv - \int {vdu}$
The first function is chosen according to the ILATE rule. So here,
I = inverse function;
L = logarithmic function;
A = algebraic function;
T = trigonometric function;
E = exponential function;
So which function comes first in this order is chosen as $u$ and the other is chosen as $v$ .
Let's take an example and solve it.
Solve $\int {x\cos (x)dx}$
Here according to ILATE rule, the algebraic function comes first
So, take $u = x{\text{ and }}v = \cos (x)$
$\Rightarrow \int {x\cos (x)dx = x\int {\cos (x)dx} - \int {\left( {\left( {\dfrac{d}{{dx}}(x)} \right)\int {\cos (x)dx} } \right)dx} }$
$\Rightarrow x(\sin x) - \int {\sin xdx}$
Remember that integration of $\cos x$ is $\sin x$ .
$\Rightarrow x(\sin x) - ( - \cos x) + c$ ----where $c$ is an arbitrary constant.
$\Rightarrow x(\sin x) + \cos x + c$
Thus we have solved a question using partial integration.

Note: After integrating, do not forget to add the arbitrary constant. The reason is, if you differentiate a constant you get zero, that means, the result that you get after integration may have a constant, which on differentiation gives you the same value, which was present in the question (before integration). And in some questions, you may also get definite integrals, for which you need to substitute the values of upper boundary and the lower boundary in the answer that you get after integrating, and find the difference between them to get the answer.