Integration by Parts
If we have studied the integral substitution method consisting of two versions, it is incomplete if we do not study the partial integral method. Because both of these methods are often used in solving integral problems both integral and uncertain integrals.
If integration uses substitution fails, we might be able to use double substitution, which is better known as partial integral. This method is based on the integration of the formula for derivatives of the product of two functions.
Suppose that $ u = u (x) $ and $ v = v (x) $, then:
$ D_x [u.v] = u.v '+ v.u' $
or
$ u.v '= D_x [u.v] -v.u' $
By integrating the two segments of the equation, we obtain
$ \int uv ' \ dx = uv - \int vu' \ dx $
Because $ dv = v ' \ dx $ and u $ du = u' \ dx $, the above equation is usually written with the symbol as follows.
Look for $ \int x \ cos \ x \ dx $
Settlement:
The question cannot indeed be done by using a substitution method. We want to write $ x \ cos \ x dx $ as $ u \ dv $. The possibility is to suppose that $ u = x $ and $ dv = cos \ x \ dx $ or $ u = cos \ x $ and $ dv = x \ dx $. We must be smart to choose between the two which are more appropriate. To determine which one has to be $ u $ and $ dv $, you can read the way in another article titled Select u and dv in Integral Partial, I can.
For example, $ u = x $ and $ dv = cos \ x \ dx $
Then,
$ du = dx $ and
$ \begin{align} v & = \int cos \ x \ dx \\ & = sin \ x \end{align} $
So that,
$ \begin{align} & \int \underbrace{x}_u \underbrace{cos \ x \ dx}_{dv} \\ & = \underbrace{x}_{u} \underbrace{sin \ x}_{v} - \int \underbrace{sin \ x}_{v} \underbrace{dx}_{du} \\ & = x \ sin \ x \ - (-cos \ x) + C \\ & = x \ sin \ x + cos \ x + C \end{align} $
If integration uses substitution fails, we might be able to use double substitution, which is better known as partial integral. This method is based on the integration of the formula for derivatives of the product of two functions.
Suppose that $ u = u (x) $ and $ v = v (x) $, then:
$ D_x [u.v] = u.v '+ v.u' $
or
$ u.v '= D_x [u.v] -v.u' $
By integrating the two segments of the equation, we obtain
$ \int uv ' \ dx = uv - \int vu' \ dx $
Because $ dv = v ' \ dx $ and u $ du = u' \ dx $, the above equation is usually written with the symbol as follows.
$ \int u \ dv = uv - \int v \ du $The formulas that correspond to the integral of course are:
$ \int_{a}^{b} u \ dv = [uv]_{a}^{b} - \int_{a}^{b} v \ du $Question example:
Look for $ \int x \ cos \ x \ dx $
Settlement:
The question cannot indeed be done by using a substitution method. We want to write $ x \ cos \ x dx $ as $ u \ dv $. The possibility is to suppose that $ u = x $ and $ dv = cos \ x \ dx $ or $ u = cos \ x $ and $ dv = x \ dx $. We must be smart to choose between the two which are more appropriate. To determine which one has to be $ u $ and $ dv $, you can read the way in another article titled Select u and dv in Integral Partial, I can.
For example, $ u = x $ and $ dv = cos \ x \ dx $
Then,
$ du = dx $ and
$ \begin{align} v & = \int cos \ x \ dx \\ & = sin \ x \end{align} $
So that,
$ \begin{align} & \int \underbrace{x}_u \underbrace{cos \ x \ dx}_{dv} \\ & = \underbrace{x}_{u} \underbrace{sin \ x}_{v} - \int \underbrace{sin \ x}_{v} \underbrace{dx}_{du} \\ & = x \ sin \ x \ - (-cos \ x) + C \\ & = x \ sin \ x + cos \ x + C \end{align} $
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