Calculus

Integration of Functions

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Integration by Substitution

  • In this topic we shall see an important method for evaluating many complicated integrals.

    Substitution for integrals corresponds to the chain rule for derivatives.

    Suppose that \(F\left( u \right)\) is an antiderivative of \(f\left( u \right):\)

    \[{\int {f\left( u \right)du} = F\left( u \right) + C.}\]

    Assuming that \(u = u\left( x \right)\) is a differentiable function and using the chain rule, we have

    \[{\frac{d}{{dx}}F\left( {u\left( x \right)} \right) }={ F^\prime\left( {u\left( x \right)} \right)u^\prime\left( x \right) }={ f\left( {u\left( x \right)} \right)u^\prime\left( x \right).}\]

    Integrating both sides gives

    \[{\int {f\left( {u\left( x \right)} \right)u^\prime\left( x \right)dx} }={ F\left( {u\left( x \right)} \right) + C.}\]

    Hence

    \[{\int {{f\left( {u\left( x \right)} \right)}{u^\prime\left( x \right)}dx} }={ \int {f\left( u \right)du},\;\;}\kern0pt{\text{where}\;\;{u = u\left( x \right)}.}\]

    This is the substitution rule formula for indefinite integrals.

    Note that the integral on the left is expressed in terms of the variable \(x.\) The integral on the right is in terms of \(u.\)

    The substitution method (also called \(u-\)substitution) is used when an integral contains some function and its derivative. In this case, we can set \(u\) equal to the function and rewrite the integral in terms of the new variable \(u.\) This makes the integral easier to solve.

    Do not forget to express the final answer in terms of the original variable \(x!\)


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Compute the integral \(\int {{e^{\large{\frac{x}{2}}\normalsize}}dx}.\)

    Example 2

    Find the integral \(\int {{{\left( {3x + 2} \right)}^5}dx}.\)

    Example 3

    Find the integral \(\int {\large{\frac{{dx}}{{\sqrt {1 + 4x} }}}\normalsize}.\)

    Example 4

    Evaluate the integral \(\int {\large{\frac{{xdx}}{{\sqrt {1 + {x^2}} }}}\normalsize}.\)

    Example 5

    Calculate the integral \(\int {\large\frac{{dx}}{{\sqrt {{a^2} – {x^2}} }}\normalsize}.\)

    Example 6

    Evaluate the integral \({\int {{\large{\frac{{{x^2}}}{{{x^3} + 1}}}\normalsize}dx} }\) using an appropriate substitution.

    Example 7

    Find the integral \(\int {\sqrt[3]{{1 – 3x}}dx}.\)

    Example 8

    Find the integral \(\int {{\large\frac{{x + 1}}{{{x^2} + 2x – 5}}\normalsize} dx}.\)

    Example 9

    Compute the integral \(\int {\large{\frac{{xdx}}{{1 + {x^4}}}}\normalsize}.\)

    Example 10

    Evaluate the integral \(\int {\large{\frac{{xdx}}{{{x^4} + 2{x^2} + 1}}}\normalsize}.\)

    Example 11

    Calculate the integral \(\int {{2^x}{e^x}dx}.\)

    Example 12

    Find the integral \(\int {x{e^{ – {x^2}}}dx}.\)

    Example 13

    Evaluate the integral \(\int {\large{\frac{{\sin x}}{{1 – \cos x}}}\normalsize dx}.\)

    Example 14

    Evaluate the integral \(\int {x\sqrt {x + 1} dx}.\)

    Example 15

    Calculate the integral \(\int {\cot \left( {3x + 5} \right)dx}.\)

    Example 16

    Find the integral \(\int {{\large\frac{{\sin 2x}}{{\sqrt {1 + {{\cos }^2}x} }}\normalsize} dx}.\)

    Example 1.

    Compute the integral \(\int {{e^{\large{\frac{x}{2}}\normalsize}}dx}.\)

    Solution.

    Let \(u = \large{\frac{x}{2}}\normalsize.\) Then

    \[{du = \frac{{dx}}{2},}\;\; \Rightarrow {dx = 2du.}\]

    So now we can easily integrate:

    \[{\int {{e^{\frac{x}{2}}}dx} = \int {{e^u} \cdot 2du} }={ 2\int {{e^u}du} }={ 2{e^u} + C }={ 2{e^{\frac{x}{2}}} + C.}\]

    Example 2.

    Find the integral \(\int {{{\left( {3x + 2} \right)}^5}dx}.\)

    Solution.

    We make the substitution \(u = 3x + 2.\) Then

    \[du = d\left( {3x + 2} \right) = 3dx.\]

    So the differential \(dx\) is given by

    \[dx = \frac{{du}}{3}.\]

    Plug all this in the integral:

    \[{\int {{{\left( {3x + 2} \right)}^5}dx} = \int {{u^5}\frac{{du}}{3}} }={ \frac{1}{3}\int {{u^5}du} }={ \frac{1}{3} \cdot \frac{{{u^6}}}{6} + C }={ \frac{{{u^6}}}{{18}} + C }={ \frac{{{{\left( {3x + 2} \right)}^6}}}{{18}} + C.}\]

    Example 3.

    Find the integral \(\int {\large{\frac{{dx}}{{\sqrt {1 + 4x} }}}\normalsize}.\)

    Solution.

    We can try to use the substitution \(u = 1 + 4x.\) Hence

    \[du = d\left( {1 + 4x} \right) = 4dx,\]

    so

    \[dx = \frac{{du}}{4}.\]

    This yields

    \[{\int {\frac{{dx}}{{\sqrt {1 + 4x} }}} = \int {\frac{{\frac{{du}}{4}}}{{\sqrt u }}} }={ \frac{1}{4}\int {\frac{{du}}{{\sqrt u }}} }={ \frac{1}{4}\int {{u^{ – \frac{1}{2}}}du} }={ \frac{1}{4} \cdot \frac{{{u^{\frac{1}{2}}}}}{{\frac{1}{2}}} + C }={ \frac{1}{4} \cdot 2{u^{\frac{1}{2}}} + C }={ \frac{{{u^{\frac{1}{2}}}}}{2} + C }={ \frac{{\sqrt u }}{2} + C }={ \frac{{\sqrt {1 + 4x} }}{2} + C.}\]

    Example 4.

    Evaluate the integral \(\int {\large{\frac{{xdx}}{{\sqrt {1 + {x^2}} }}}\normalsize}.\)

    Solution.

    Let \(u = 1 + {x^2}.\) Then

    \[du = d\left( {1 + {x^2}} \right) = 2xdx.\]

    We see that

    \[xdx = \frac{{du}}{2}.\]

    Hence

    \[{\int {\frac{{xdx}}{{\sqrt {1 + {x^2}} }}} }={ \int {\frac{{\frac{{du}}{2}}}{{\sqrt u }}} }={ \int {\frac{{du}}{{2\sqrt u }}} }={ \sqrt u + C }={ \sqrt {1 + {x^2}} + C.}\]

    Example 5.

    Calculate the integral \(\int {\large\frac{{dx}}{{\sqrt {{a^2} – {x^2}} }}\normalsize}.\)

    Solution.

    Let \(u = \large\frac{x}{a}\normalsize.\) Then \(x = au,\) \(dx = adu.\) Hence, the integral is

    \[\require{cancel}
    {\int {\frac{{dx}}{{\sqrt {{a^2} – {x^2}} }}} }
    = {\int {\frac{{adu}}{{\sqrt {{a^2} – {{\left( {au} \right)}^2}} }}} }
    = {\int {\frac{{adu}}{{\sqrt {{a^2}\left( {1 – {u^2}} \right)} }}} }
    = {\int {\frac{{\cancel{a}du}}{{\cancel{a}\sqrt {1 – {u^2}} }}} }
    = {\int {\frac{{du}}{{\sqrt {1 – {u^2}} }}} }
    = {\arcsin u + C }
    = {\arcsin \frac{x}{a} + C.}
    \]

    Example 6.

    Evaluate the integral \({\int {{\large{\frac{{{x^2}}}{{{x^3} + 1}}}\normalsize}dx} }\) using an appropriate substitution.

    Solution.

    We try the substitution \(u = {x^3} + 1.\)

    Calculate the differential \(du:\)

    \[{du = d\left( {{x^3} + 1} \right) = 3{x^2}dx.}\]

    We see from the last expression that

    \[{{x^2}dx = \frac{{du}}{3},}\]

    so we can rewrite the integral in terms of the new variable \(u:\)

    \[{I = \int {\frac{{{x^2}}}{{{x^3} + 1}}dx} }={ \int {\frac{{\frac{{du}}{3}}}{u}} }={ \int {\frac{{du}}{{3u}}} .}\]

    Now we can easily evaluate this integral:

    \[{I = \int {\frac{{du}}{{3u}}} }={ \frac{1}{3}\int {\frac{{du}}{u}} }={{\frac{1}{3}\ln \left| u \right|} + C.}\]

    Express the result in terms of the variable \(x:\)

    \[{I = \frac{1}{3}\ln \left| u \right| + C }={{ \frac{1}{3}\ln \left| {{x^3} + 1} \right| + C}}.\]

    Page 1
    Problems 1-6
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    Problems 7-16