Integration by Substitution

In this tutorial we shall use different substitutions before evaluating the integrals. Sometimes integration are very complicated to solve so in this situation we shall see that a suitable substitution will make the integral simpler one and then it will be easy to tackle it.

Most of integration formulae and problems are also proven and solved by this method of substitution, and we shall inspect some of them by 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 result putting x = a\sin t, by taking differentiation we have dx = a\cos tdt, also we 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 result putting x = a\tan t, by taking differentiation we have dx = a{\sec ^2}tdt, also we 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}