Integration by Parts
Solved Problems
Example 9.
Compute the integral \[\int {{x^2}{e^x}dx}.\]
Solution.
Let
\[u = {x^2},\;\;dv = {e^x}dx.\]
Then
\[du = 2xdx,\;\;v = \int {{e^x}dx} = {e^x},\]
The integral is written as
\[\int {{x^2}{e^x}dx} = {x^2}{e^x} - \int {2x{e^x}dx} = {x^2}{e^x} - 2\int {x{e^x}dx} .\]
We calculate the last integral by repeated integration by parts. Choosing
\[u = x,\;\;dv = {e^x}dx,\]
we obtain
\[du = dx,\;\;v = \int {{e^x}dx} = {e^x},\]
so
\[\int {{x^2}{e^x}dx} = {x^2}{e^x} - 2\int {x{e^x}dx} = {x^2}{e^x} - 2\left( {x{e^x} - \int {{e^x}dx} } \right) = {x^2}{e^x} - 2\left( {x{e^x} - {e^x}} \right) + C = {x^2}{e^x} - 2x{e^x} + 2{e^x} + C = {e^x}\left( {{x^2} - 2x + 2} \right) + C.\]
Example 10.
Calculate the integral \[\int {\arcsin xdx}.\]
Solution.
Use integration by parts: \(u = \arcsin x,\) \(dv = dx.\) Then
\[du = \frac{d}{{dx}}\arcsin x = \frac{{dx}}{{\sqrt {1 - {x^2}} }}, v = \int {dx} = x,\]
and the integral becomes
\[\int {\arcsin xdx} = x\arcsin x - \int {\frac{x}{{\sqrt {1 - {x^2}} }}dx}.\]
To calculate the new integral, we substitute \(w = 1 - {x^2}.\) In this case,
\[dw = - 2xdx, xdx = - \frac{{dw}}{2},\]
so that the integral in the right side is
\[\int {\frac{x}{{\sqrt {1 - {x^2}} }}dx} = \int {\frac{{\left( { - \frac{{dw}}{2}} \right)}}{{\sqrt w }}} = - \int {\frac{{dw}}{{2\sqrt w }}} = - \sqrt w + C = - \sqrt {1 - {x^2}} + C.\]
The result is
\[\int {\arcsin xdx} = x\arcsin x - \left( { - \sqrt {1 - {x^2}} } \right) + C = x\arcsin x + \sqrt {1 - {x^2}} + C.\]
Example 11.
Find the integral \[\int {\arctan xdx}.\]
Solution.
According to the ILATE rule, let
\[u = \arctan x,\;\;dv = dx.\]
Then
\[du = \frac{{dx}}{{1 + {x^2}}},\;\;v = x,\]
and
\[I = \int {\arctan xdx} = x\arctan x - \int {\frac{{xdx}}{{1 + {x^2}}}} .\]
We use integration by substitution to find the last integral:
\[t = {x^2} + 1,\;\; \Rightarrow dt = 2xdx,\;\; \Rightarrow xdx = \frac{{dt}}{2},\]
so
\[\int {\frac{{xdx}}{{1 + {x^2}}}} = \int {\frac{{\frac{{dt}}{2}}}{t}} = \frac{1}{2}\int {\frac{{dt}}{t}} = \frac{1}{2}\ln \left| t \right| = \frac{1}{2}\ln \left| {1 + {x^2}} \right| = \frac{1}{2}\ln \left( {1 + {x^2}} \right).\]
Hence, the original integral is given by
\[I = x\arctan x - \frac{1}{2}\ln \left( {1 + {x^2}} \right) + C.\]
Example 12.
Find the integral \[\int {{e^x}\sin xdx}.\]
Solution.
We use integration by parts:
\[\int {udv} = uv - \int {vdu} .\]
Let \(u = {e^x},\) \(dv = \sin xdx.\) Then
\[du = {e^x}dx, v = \int {\sin xdx} = - \cos x,\]
so the integral can be written in the form:
\[\int {{e^x}\sin xdx} = - {e^x}\cos x + \int {{e^x}\cos xdx} .\]
Apply integration by parts one more time. Now let \(u = {e^x},\) \(dv = \cos xdx.\) Hence,
\[du = {e^x}dx, v = \int {\cos xdx} = \sin x,\]
so that the integral we started with is
\[\int {{e^x}\sin xdx} = - {e^x}\cos x + \int {{e^x}\cos xdx} = - {e^x}\cos x + {e^x}\sin x - \int {{e^x}\sin xdx}.\]
Solving this equation for this integral, we obtain
\[2\int {{e^x}\sin xdx} = {e^x}\sin x - {e^x}\cos x\;\;\;\text{or}\;\;\;\int {{e^x}\sin xdx} = \frac{{{e^x}\left( {\sin x - \cos x} \right)}}{2} + C.\]
Example 13.
Find the integral \[\int {{e^{ - x}}\sin xdx}.\]
Solution.
We use integration by parts. Let
\[u = \sin x,\;\;dv = {e^{ - x}}dx.\]
Then
\[du = \cos xdx,\;\;v = \int {{e^{ - x}}dx} = - {e^{ - x}},\]
so that the integral is written as
\[\int {{e^{ - x}}\sin xdx} = - {e^{ - x}}\sin x - \int {\left( { - {e^{ - x}}} \right)\cos xdx} = - {e^{ - x}}\sin x + \int {{e^{ - x}}\cos xdx} .\]
The new integral in the right-hand side can be evaluated by repeated integration by parts. We choose
\[u = \cos x,\;\;dv = {e^{ - x}}dx.\]
Hence
\[du = - \sin xdx,\;\;v = \int {{e^{ - x}}dx} = - {e^{ - x}}.\]
Then the original integral is expressed in the form
\[\int {{e^{ - x}}\sin xdx} = - {e^{ - x}}\sin x + \int {{e^{ - x}}\cos xdx} = - {e^{ - x}}\sin x + \Bigl[ { - {e^{ - x}}\cos x - \int {\left( { - {e^{ - x}}} \right)\left( { - \sin x} \right)dx} } \Bigr] = - {e^{ - x}}\sin x - {e^{ - x}}\cos x - \int {{e^{ - x}}\sin xdx} .\]
We obtain the equation for our integral \(\int {{e^{ - x}}\sin xdx} .\) Solving it, we find that
\[\int {{e^{ - x}}\sin xdx} = - \frac{{{e^{ - x}}}}{2}\left( {\sin x + \cos x} \right).\]
Example 14.
Compute the integral \[\int {{{\sin }^2}xdx}.\]
Solution.
Let
\[u = {\sin ^2}x,\;\;dv = dx.\]
Then
\[du = 2\sin x\cos xdx = \sin 2xdx,\;\;v = \int {dx} = x.\]
Integrating by parts, we get
\[I = \int {{{\sin }^2}xdx} = x\,{\sin ^2}x - \int {x\sin 2xdx} .\]
We evaluate the integral in the right-hand side by repeated integration by parts. Choosing
\[u = x,\;\;dv = \sin 2xdx,\]
we have
\[du = dx,\;\;v = \int {\sin 2xdx} = - \frac{1}{2}\cos 2x.\]
So, the last integral is given by
\[\int {x\sin 2xdx} = - \frac{x}{2}\cos 2x - \int {\left( { - \frac{1}{2}\cos 2x} \right)dx} = - \frac{x}{2}\cos 2x + \frac{1}{2}\int {\cos 2xdx} = - \frac{x}{2}\cos 2x + \frac{1}{2} \cdot \frac{1}{2}\sin 2x + C = - \frac{x}{2}\cos 2x + \frac{1}{4}\sin 2x + C.\]
Hence, the initial integral is written as
\[I = x\,{\sin ^2}x - \int {x\sin 2xdx} = x\,{\sin ^2}x + \frac{x}{2}\cos 2x - \frac{1}{4}\sin 2x + C.\]
We can simplify this expression using the trig identities
\[\cos 2x = {\cos ^2}x - {\sin ^2}x,\;\;{\sin ^2}x + {\cos^2}x = 1.\]
This yields:
\[I = x\,{\sin ^2}x + \frac{x}{2}\left( {{{\cos }^2}x - {{\sin }^2}x} \right) - \frac{1}{4}\sin 2x + C = x\,{\sin ^2}x + \frac{x}{2}{\cos ^2}x - \frac{x}{2}{\sin ^2}x - \frac{1}{4}\sin 2x + C = \frac{x}{2}\left( {{{\sin }^2}x + {{\cos }^2}x} \right) - \frac{1}{4}\sin 2x + C = \frac{x}{2} - \frac{1}{4}\sin 2x + C.\]
Example 15.
Compute the integral \[\int {{{\cos }^2}xdx}.\]
Solution.
We use integration by part. So let
\[u = {\cos ^2}x,\;\;dv = dx.\]
Then
\[du = -2\cos x\sin xdx = -\sin 2xdx,\;\;v = \int {dx} = x.\]
The integral becomes equal
\[I = \int {{{\cos }^2}xdx} = x\,{\cos ^2}x - \int {x\left( { - \sin 2x} \right)dx} = x\,{\cos ^2}x + \int {x\sin 2xdx} .\]
The last integral can be evaluated by repeated integration by parts. Let
\[u = x,\;\;dv = \sin 2xdx.\]
Hence
\[du = dx,\;\;v = \int {\sin 2xdx} = - \frac{1}{2}\cos 2x,\]
and
\[\int {x\sin 2xdx} = - \frac{x}{2}\cos 2x - \int {\left( { - \frac{1}{2}\cos 2x} \right)dx} = - \frac{x}{2}\cos 2x + \frac{1}{2}\int {\cos 2xdx} = - \frac{x}{2}\cos 2x + \frac{1}{2} \cdot \frac{1}{2}\sin 2x + C = - \frac{x}{2}\cos 2x + \frac{1}{4}\sin 2x + C.\]
Consequently, the initial integral can be written in the form
\[I = x\,{\cos^2}x + \int {x\sin 2xdx} = x\,{\cos^2}x - \frac{x}{2}\cos 2x + \frac{1}{4}\sin 2x + C.\]
We can simplify this expression using the trig identities
\[\cos 2x = {\cos ^2}x - {\sin ^2}x,\;\;{\sin ^2}x + {\cos^2}x = 1.\]
This gives:
\[I = x\,{\cos^2}x - \frac{x}{2}\left( {{{\cos }^2}x - {{\sin }^2}x} \right) + \frac{1}{4}\sin 2x + C = x\,{\cos^2}x - \frac{x}{2}{\cos ^2}x + \frac{x}{2}{\sin ^2}x + \frac{1}{4}\sin 2x + C = \frac{x}{2}\left( {{{\cos }^2}x + {{\sin }^2}x} \right) + \frac{1}{4}\sin 2x + C = \frac{x}{2} + \frac{1}{4}\sin 2x + C.\]
Example 16.
Find a reduction formula for \[\int {{{\sin }^n}xdx} , n \ge 2.\]
Solution.
We apply integration by parts: \(\int {udv} \) \(= uv - \int {vdu}.\) Let \(u = {\sin ^{n - 1}}x,\) \(dv = \sin xdx.\) Then
\[\int {udv} = uv - \int {vdu}.\]
Let \(u = {\sin ^{n - 1}}x,\) \(dv = \sin xdx.\) Then
\[du = \frac{d}{{dx}}{\sin ^{n - 1}}x = \left( {n - 1} \right){\sin ^{n - 2}}x\cos xdx,\;\;\;v = \int {\sin xdx} = - \cos x.\]
Hence,
\[\int {{{\sin }^n}xdx} = - \cos x\,{\sin ^{n - 1}}x - \int {\left( { - \cos x} \right)\left( {n - 1} \right)\,{{\sin }^{n - 2}}x\cos x}dx = - \cos x\,{\sin ^{n - 1}}x - \int {\left( {n - 1} \right)\,{{\sin }^{n - 2}}x\,{{\cos }^2}xdx} = - \cos x\,{\sin ^{n - 1}}x + \int {\left( {n - 1} \right)\,{{\sin }^{n - 2}}x\left( {1 - {{\sin }^2}x} \right)dx} = - \cos x\,{\sin ^{n - 1}}x + \left( {n - 1} \right)\int {{{\sin }^{n - 2}}xdx} - \left( {n - 1} \right)\int {{{\sin }^n}xdx} .\]
Solve the equation for \(\int {{{\sin }^n}xdx}.\) As a result, we obtain
\[\int {{{\sin }^n}xdx} + \left( {n - 1} \right)\int {{{\sin }^n}xdx} = - \cos x\,{\sin ^{n - 1}}x + \left( {n - 1} \right)\int {{{\sin }^{n - 2}}xdx} ,\;\; \Rightarrow n\int {{{\sin }^n}xdx} = \left( {n - 1} \right)\int {{{\sin }^{n - 2}}xdx} - \cos x\,{\sin ^{n - 1}}x,\;\; \Rightarrow \int {{{\sin }^n}xdx} = \frac{{n - 1}}{n}\int {{{\sin }^{n - 2}}xdx} - \frac{{\cos x\,{{\sin }^{n - 1}}x}}{n}.\]
