Calculus

Integration of Functions

Change of Variable

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

Let \(F\left( x \right)\) be an indefinite integral or antiderivative of \(f\left( x \right).\) Then
\[
{\int {f\left( x \right)dx} }
= {\int {f\left( {g\left( u \right)} \right)g’\left( u \right)du} }
= {F\left( u \right) }
= {F\left( {{g^{ – 1}}\left( x \right)} \right),}
\] where \(x = g\left( u \right)\) is a substitution. Accordingly, the inverse function \(u = {g^{ – 1}}\left( x \right)\) describes the dependence of the new variable on the old variable.

It’s important to remember that the differential \(dx\) also needs to be substituted. It must be replaced with the differential of the new variable \(du.\) For definite integrals, it is also necessary to change the limits of integration. See about this on the page “The Definite Integral and Fundamental Theorem of Calculus“.

Solved Problems

Click on problem description to see 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