Integration by Substitution
Solved Problems
Example 9.
Compute the integral \[\int {\frac{{xdx}}{{1 + {x^4}}}}.\]
Solution.
We can try the substitution \(u = {x^2}.\) Then
\[du = 2xdx,\;\; \Rightarrow xdx = \frac{{du}}{2}.\]
Hence, the integral is equal to
\[\int {\frac{{xdx}}{{1 + {x^4}}}} = \int {\frac{{\frac{{du}}{2}}}{{1 + {u^2}}}} = \frac{1}{2}\int {\frac{{du}}{{1 + {u^2}}}} = \frac{1}{2}\arctan u + C = \frac{1}{2}\arctan {x^2} + C.\]
Example 10.
Evaluate the integral \[\int {\frac{{xdx}}{{{x^4} + 2{x^2} + 1}}}.\]
Solution.
If we complete the square in the denominator:
\[{x^4} + 2{x^2} + 1 = {\left( {{x^2} + 1} \right)^2} = {\left( {1 + {x^2}} \right)^2},\]
we can use the substitution \(u = 1 + {x^2}.\) Write the differential \(du:\)
\[du = d\left( {1 + {x^2}} \right) = 2xdx.\]
We see that the numerator can be expressed in terms of \(u:\)
\[xdx = \frac{{du}}{2}.\]
Hence, the integral is given by
\[\int {\frac{{xdx}}{{{x^4} + 2{x^2} + 1}}} = \int {\frac{{xdx}}{{{{\left( {1 + {x^2}} \right)}^2}}}} = \int {\frac{{\frac{{du}}{2}}}{{{u^2}}}} = \frac{1}{2}\int {\frac{{du}}{{{u^2}}}} = \frac{1}{2}\int {{u^{ - 2}}du} = \frac{1}{2} \cdot \frac{{{u^{ - 1}}}}{{\left( { - 1} \right)}} + C = - \frac{1}{{2u}} + C = - \frac{1}{{2\left( {1 + {x^2}} \right)}} + C.\]
Example 11.
Calculate the integral \[\int {{2^x}{e^x}dx}.\]
Solution.
We rewrite the integral in the following way:
\[\int {{2^x}{e^x}dx} = \int {{{\left( {2e} \right)}^x}dx} .\]
Denoting \(2e = a\) (this is not a change of variable, since \(x\) still remains the independent variable), we get the table integral:
\[\int {{{\left( {2e} \right)}^x}dx} = \int {{a^x}dx} = \frac{{{a^x}}}{{\ln a}} + C = \frac{{{{\left( {2e} \right)}^x}}}{{\ln \left( {2e} \right)}} + C = \frac{{{2^x}{e^x}}}{{\ln 2 + \ln e}} + C = \frac{{{2^x}{e^x}}}{{\ln 2 + 1}} + C.\]
Example 12.
Find the integral \[\int {x{e^{ - {x^2}}}dx}.\]
Solution.
Using the substitution \(u = - {x^2},\) we have
\[du = d\left( { - {x^2}} \right) = - 2xdx.\]
Note that
\[xdx = - \frac{{du}}{2},\]
so we can rewrite the integral in terms of the variable \(u\) and solve it:
\[\int {x{e^{ - {x^2}}}dx} = \int {{e^u}\left( { - \frac{{du}}{2}} \right)} = - \frac{1}{2}\int {{e^u}du} = - \frac{1}{2}{e^u} + C = - \frac{{{e^{ - {x^2}}}}}{2} + C.\]
Example 13.
Evaluate the integral \[\int {\frac{{\sin x}}{{1 - \cos x}} dx}.\]
Solution.
We make the substitution \(u = 1 - \cos x.\) Hence
\[du = - \left( { - \sin x} \right)dx = \sin xdx.\]
This gives
\[\int {\frac{{\sin x}}{{1 - \cos x}}dx} = \int {\frac{{du}}{u}} = \ln \left| u \right| + C = \ln \left| {1 - \cos x} \right| + C.\]
Example 14.
Evaluate the integral \[\int {x\sqrt {x + 1} dx}.\]
Solution.
To get rid of the square root, we make the substitution \(u = \sqrt {x + 1} .\) Then
\[{u^2} = x + 1,\;\; \Rightarrow x = {u^2} - 1,\;\; \Rightarrow du = 2udu.\]
The integral becomes
\[\int {x\sqrt {x + 1} dx} = \int {\left( {{u^2} - 1} \right)u \cdot 2udu} = 2\int {\left( {{u^2} - 1} \right){u^2}du} = 2\int {\left( {{u^4} - {u^2}} \right)du} = 2\int {{u^4}du} - 2\int {{u^2}du} = 2 \cdot \frac{{{u^5}}}{5} - 2 \cdot \frac{{{u^3}}}{3} + C = \frac{2}{5}{\left( {x + 1} \right)^{\frac{5}{2}}} - \frac{2}{3}{\left( {x + 1} \right)^{\frac{3}{2}}} + C.\]
Example 15.
Calculate the integral \[\int {\cot \left( {3x + 5} \right)dx}.\]
Solution.
We can write the integral as
\[\int {\cot \left( {3x + 5} \right)dx} = \int {\frac{{\cos \left( {3x + 5} \right)}}{{\sin\left( {3x + 5} \right)}}dx} .\]
Changing the variable
\[u = \sin\left( {3x + 5} \right),\;\;\;du = 3\cos \left( {3x + 5} \right)dx,\;\; \Rightarrow \cos \left( {3x + 5} \right)dx = \frac{{du}}{3},\]
we obtain the answer
\[\int {\cot \left( {3x + 5} \right)dx} = \int {\frac{{\cos \left( {3x + 5} \right)}}{{\sin\left( {3x + 5} \right)}}dx} = \int {\frac{{\frac{{du}}{3}}}{u}} = \frac{1}{3}\int {\frac{{du}}{u}} = \frac{1}{3}\ln \left| u \right| + C = \frac{1}{3}\ln \left| {\sin\left( {3x + 5} \right)} \right| + C.\]
Example 16.
Find the integral \[\int {{\frac{{\sin 2x}}{{\sqrt {1 + {{\cos }^2}x} }}} dx}.\]
Solution.
We make the following substitution:
\[u = 1 + {\cos ^2}x,\;\; \Rightarrow du = {\left( {1 + {{\cos }^2}x} \right)^\prime }dx = 2\cos x \cdot \left( { - \sin x} \right)dx = - \sin 2xdx.\]
Hence,
\[\int {\frac{{\sin 2x}}{{\sqrt {1 + {{\cos }^2}x} }}dx} = \int {\frac{{\left( { - du} \right)}}{{\sqrt u }}} = - 2\int {\frac{{du}}{{2\sqrt u }}} = - 2\sqrt u + C = - 2\sqrt {1 + {{\cos }^2}x} + C.\]
