Integration by Parts
Integration by parts method is used when we want to integrate the product of two functions. When finding the derivative of the product of two functions we use the product rule, and since the integral is the reverse of derivative then integration by parts is the reverse of the product rule.
Let’s explain this step by step
If we have two differentiable functions $ \displaystyle u(x)$ and $ \displaystyle v(x)$ and we want to find the derivative of their product we use the product rule:
$\displaystyle (u\cdot v{)}’=u\cdot {v}’+v\cdot {u}’$ or written like $ \displaystyle d(uv)=udv+vdu$
By rearranging this rule we can write:
$ \displaystyle udv=d(uv)-vdu$
By integrating both sides we get:
$ \displaystyle \int{{udv=\int{{d(uv)-\int{{vdu}}}}}}$
Simplifying, we get the integration by part formula:
$ \displaystyle \int{{udv=uv-\int{{vdu}}}}$
This formula helps us to turn a complicated integral into a simple one. The only thing we should be careful and choose wisely is ‘u’ and ‘dv’.
When should you choose ‘u’?
There are some common expressions when you should always choose that as ‘u’.
Inverse trigonometric Functions:
$ \displaystyle \arcsin x,\arccos x,\arctan x………$
Logarithmic Functions:
$ \displaystyle \log x,\ln x,{{\log }_{2}}2x,….{{\log }_{a}}nx$
Algebraic Functions:
$ \displaystyle x,{{x}^{2}},{{x}^{3}},5{{x}^{4}},….,a{{x}^{n}}$
Trigonometric Functions:
$ \displaystyle \sin x,\cos x,\tan x……..\sin nx,\cos nx$
Exponential Functions:
$ \displaystyle {{e}^{x}},{{e}^{{3x}}}{{,3}^{{-2x}}},……,{{e}^{{nx}}},{{a}^{{nx}}}$
This list is written from the top to the least priority.
Example 1: Find the integral $ \displaystyle \int{x}{{e}^{{2x}}}dx$?
In this kind of expression we should always remember that the power of x has a priority over the exponential expression.
So let’s note $ \displaystyle u=x$ than $ \displaystyle du=dx$
What’s left would be dv, so $ \displaystyle dv={{e}^{{2x}}}dx$
Integrating it we obtain v:
$ \displaystyle \int{{dv}}=\int{{{{e}^{{2x}}}}}dx$
$ \displaystyle v=\frac{1}{2}{{e}^{{2x}}}$
Now we substitute them into the integration by parts formula:
$ \displaystyle \int{{udv=uv-\int{{vdu}}}}$
$ \displaystyle \int{x}{{e}^{{2x}}}dx=x\cdot \frac{1}{2}{{e}^{{2x}}}-\int{{\frac{1}{2}{{e}^{{2x}}}dx}}$
Then try to find the integral:
$ \displaystyle \int{x}{{e}^{{2x}}}dx=\frac{1}{2}x{{e}^{{2x}}}-\frac{1}{2}\int{{{{e}^{{2x}}}dx}}$
$ \displaystyle \int{x}{{e}^{{2x}}}dx=\frac{1}{2}x{{e}^{{2x}}}-\frac{1}{4}{{e}^{{2x}}}+C$
$ \displaystyle \int{x}{{e}^{{2x}}}dx=\frac{1}{2}{{e}^{{2x}}}(x-\frac{1}{2})+C$
Note! When choosing u we should choose the one that makes du/dx on a simpler form than u.
Example 2: Find the integral $ \displaystyle \int{{3x\sin 2xdx}}$?
So let’s note $ \displaystyle u=3x$ than $ \displaystyle du=3dx$
What’s left would be dv, so $ \displaystyle dv=\sin 2xdx$
Integrating it we obtain v:
$ \displaystyle \int{{dv=\int{{\sin 2xdx}}}}$
$ \displaystyle v=-\frac{1}{2}\cos 2x$
Now we substitute them into the integration by parts formula:
$ \displaystyle \int{{udv=uv-\int{{vdu}}}}$
$ \displaystyle \int{{3x\sin 2xdx}}=3x\cdot -\frac{1}{2}\cos 2x-\int{{-\frac{1}{2}\cos 2x\cdot 3dx}}$
Then try to find the integral:
$ \displaystyle \int{{3x\sin 2xdx}}=-\frac{{3x}}{2}\cos 2x+\frac{3}{2}\int{{\cos 2xdx}}$
$ \displaystyle \int{{3x\sin 2xdx}}=-\frac{{3x}}{2}\cos 2x+\frac{3}{4}\sin 2x$
Example 3: Find the integral $ \displaystyle \int{{x\ln 2xdx}}$?
From the priorities above it’s obvious whose gone be the u and dv.
So let’s note $ \displaystyle u=\ln 2x$ than$ \displaystyle du=\frac{2}{x}dx$
What’s left would be dv, so $ \displaystyle dv=xdx$
Integrating it we obtain v:
$ \displaystyle \int{{dv=\int{{xdx}}}}$
$ \displaystyle v=\frac{{{{x}^{2}}}}{2}$
Now we substitute them into the integration by parts formula:
$ \displaystyle \int{{udv=uv-\int{{vdu}}}}$
$ \displaystyle \int{{x\ln 2xdx=\ln 2x}}\cdot \frac{{{{x}^{2}}}}{2}-\int{{\frac{{{{x}^{2}}}}{2}\cdot \frac{2}{x}dx}}$
Then try to find the integral:
$ \displaystyle \int{{x\ln 2xdx=\frac{{{{x}^{2}}\ln 2x}}{2}}}-\int{{xdx}}$
$ \displaystyle \int{{x\ln 2xdx=\frac{{{{x}^{2}}\ln 2x}}{2}}}-\frac{{{{x}^{2}}}}{2}+C$
$ \displaystyle \int{{x\ln 2xdx=\frac{{{{x}^{2}}}}{2}}}(\ln x-1)+C$
Example 4: Find the integral $\displaystyle \int{{2\text{arctgxdx}}}$
So let’s note $ \displaystyle u=\arctan x$ than $ \displaystyle du=\frac{1}{{1+{{x}^{2}}}}dx$
What’s left would be dv, so $ \displaystyle dv=2dx$
Integrating it we obtain v:
$ \displaystyle \int{{dv=\int{{2dx}}}}$
$ \displaystyle v=2x$
Now we substitute them into the integration by parts formula:
$ \displaystyle \int{{udv=uv-\int{{vdu}}}}$
$\displaystyle \int{{2\text{arctgxdx}=}}2x\arctan x-\int{{\frac{{2x}}{{1+{{x}^{2}}}}}}dx$
Then try to find the integral:
We use the u-substitution method to find the integral $ \displaystyle \int{{\frac{{2x}}{{1+{{x}^{2}}}}}}dx$
We substitute $ \displaystyle u=1+{{x}^{2}}$ than $ \displaystyle du=2xdx$
Finding the integral by substituting:
$ \displaystyle \int{{\frac{{2x}}{{1+{{x}^{2}}}}}}dx=\int{{\frac{{du}}{u}}}$
$ \displaystyle \int{{\frac{{du}}{u}}}=\ln u+C$
Now we substitute back the $ \displaystyle u=1+{{x}^{2}}$
$ \displaystyle \int{{\frac{{2x}}{{1+{{x}^{2}}}}}}dx=\ln (1+{{x}^{2}})+C$
Example 5: Find the integral$ \displaystyle \int{{{{2}^{x}}\cdot {{x}^{2}}}}dx$?
Based on the priorities of the table:
So let’s note $ \displaystyle u={{x}^{2}}$ than$ \displaystyle du=2xdx$
What’s left would be dv, so $ \displaystyle dv={{2}^{x}}dx$
Integrating it we obtain v:
$ \displaystyle \int{{dv=\int{{{{2}^{x}}dx}}}}$
$ \displaystyle v=\frac{{{{2}^{x}}}}{{\ln 2}}$
Now we substitute them into the integration by parts formula:
$ \displaystyle \int{{udv=uv-\int{{vdu}}}}$
$ \displaystyle \int{{{{2}^{x}}\cdot {{x}^{2}}}}dx={{x}^{2}}\cdot \frac{{{{2}^{x}}}}{{\ln 2}}-\int{{\frac{{{{2}^{x}}}}{{\ln 2}}\cdot }}2xdx$
Then try to find the integral:
$ \displaystyle \int{{{{2}^{x}}\cdot {{x}^{2}}}}dx=\frac{{{{2}^{x}}{{x}^{2}}}}{{\ln 2}}-\frac{2}{{\ln 2}}\int{{x\cdot {{2}^{x}}}}dx$
For the integral we are going to use again the integration by parts:
$ \displaystyle \int{{x\cdot {{2}^{x}}}}dx=x\cdot \frac{{{{2}^{x}}}}{{\ln 2}}-\int{{\frac{{{{2}^{x}}}}{{\ln 2}}dx}}$
$ \displaystyle \int{{x\cdot {{2}^{x}}}}dx=\frac{{{{2}^{x}}x}}{{\ln 2}}-\frac{{{{2}^{x}}}}{{{{{\ln }}^{2}}2}}$
Now we obtain the integral:
$ \displaystyle \int{{{{2}^{x}}\cdot {{x}^{2}}}}dx=\frac{{{{2}^{x}}{{x}^{2}}}}{{\ln 2}}-\frac{2}{{\ln 2}}(\frac{{{{2}^{x}}x}}{{\ln 2}}-\frac{{{{2}^{x}}}}{{{{{\ln }}^{2}}2}})+C$
$ \displaystyle \int{{{{2}^{x}}\cdot {{x}^{2}}}}dx=\frac{{{{2}^{x}}{{x}^{2}}}}{{\ln 2}}-\frac{{{{2}^{x}}^{{+1}}x}}{{{{{\ln }}^{2}}2}}+\frac{{{{2}^{x}}^{{+1}}}}{{{{{\ln }}^{3}}2}}+C$