Integration by Substitution

In this tutorial we shall use different substitutions before evaluating the integrals. Sometimes an integration is very complicated to solve, so in this situation we shall see that a suitable substitution will make the integral simpler, making it easier to tackle.

Most integration formulae and problems are also proven and solved by this method of substitution, and we shall look at some of them using this method.

Example: Evaluate the integral

\int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = {{\sin }^{ - 1}}\left( {\frac{x}{a}} \right) + c}

To solve this put x = a\sin t, and with differentiation we have dx = a\cos tdt. We also have t = {\sin ^{ - 1}}\left( {\frac{x}{a}} \right)

So the given integral takes the form

\begin{gathered} \int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = \int {\frac{{a\cos t}}{{\sqrt {{a^2} - {a^2}{{\sin }^2}t} }}dt} } \\ \Rightarrow \int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = \int {\frac{{a\cos t}}{{a\sqrt {1 - {{\sin }^2}t} }}dt} } = \int {\frac{{\cos t}}{{\sqrt {{{\cos }^2}t} }}dt} \\ \Rightarrow \int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = \int {\frac{{\cos t}}{{\cos t}}dt = \int {1dt} } } \\ \Rightarrow \int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = t + c} \\ \Rightarrow \int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = {{\sin }^{ - 1}}\left( {\frac{x}{a}} \right) + c} \\ \end{gathered}

Example: Evaluate the integral

\int {\frac{{dx}}{{{a^2} + {x^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right) + c}

To solve this put x = a\tan t, and with differentiation we have dx = a{\sec ^2}tdt. We also have t = {\tan ^{ - 1}}\left( {\frac{x}{a}} \right)

So the given integral takes the form

\begin{gathered} \int {\frac{{dx}}{{{a^2} + {x^2}}} = \int {\frac{{a{{\sec }^2}t}}{{{a^2} + {a^2}{{\tan }^2}t}}dt = \int {\frac{{a{{\sec }^2}t}}{{{a^2}\left( {1 + {{\tan }^2}t} \right)}}dt} } } \\ \Rightarrow \int {\frac{{dx}}{{{a^2} + {x^2}}} = \frac{1}{a}\int {\frac{{{{\sec }^2}t}}{{\left( {1 + {{\tan }^2}t} \right)}}dt} } = \frac{1}{a}\int {dt} \\ \Rightarrow \int {\frac{{dx}}{{{a^2} + {x^2}}} = \frac{1}{a}t + c} \\ \Rightarrow \int {\frac{{dx}}{{{a^2} + {x^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right) + c} \\ \end{gathered}