Calculus

Integration of Functions

Integration of Functions Logo

Change of Variable

  • Let \(F\left( x \right)\) be an indefinite integral or antiderivative of \(f\left( x \right).\) To evaluate an integral, it is often necessary to make a substitution. Suppose that

    \[
    {\int {f\left( x \right)dx} }
    = {\int {f\left( {g\left( u \right)} \right)g’\left( u \right)du} }
    = {F\left( u \right) }
    = {F\left( {{g^{ – 1}}\left( x \right)} \right),}
    \]

    where \(x = g\left( u \right)\) is the new substitution. Respectively, the inverse function \(u = {g^{ – 1}}\left( x \right)\) determines the new variable depending on the old one.

    Making a substitution, we should also replace the differential in the old variable \(dx\) with the new differential \(du.\) For a definite integral, we should also change the limits of integration.


    Solved Problems

    Click a problem to see the solution.

    Example 1

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

    Example 2

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

    Example 3

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

    Example 4

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

    Example 5

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

    Example 1.

    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 2.

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

    Solution.

    We make the substitution \(u = {x^2} + 2x – 5.\) Then \(du = 2xdx + 2dx \) \(= 2\left( {x + 1} \right)dx\) or \(\left( {x + 1} \right)dx = {\large\frac{{du}}{2}\normalsize}.\) The integral is easy to calculate with the new variable:

    \[
    {\int {\frac{{x + 1}}{{{x^2} + 2x – 5}}dx} }
    = {\int {\frac{{\frac{{du}}{2}}}{u}} }
    = {\frac{1}{2}\int {\frac{{du}}{u}} }
    = {\frac{1}{2}\ln \left| u \right| + C }
    = {\frac{1}{2}\ln \left| {{x^2} + 2x – 5} \right| }+{ C.}
    \]

    Page 1
    Problems 1-2
    Page 2
    Problems 3-5