Integration by Substitution | Summary & Examples

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1 Summary

1.1 The Substitution Rule is defined as:

1.2 Example #1

1.3 Example #2

1.3.1 Reference

Summary

Substitution Rule is defined as:

Indefinite Integral:

$\int { f\left( g\left( x \right) \right) } g^{ ' }\left( x \right) dx\quad =\quad \int { f\left( u \right) } du$

Definite Integral:

$\int _{ a }^{ b }{ f\left( g\left( x \right) \right) g^{ ' }\left( x \right) dx } \quad =\quad \int _{ g\left( a \right) }^{ g\left( b \right) }{ f\left( u \right) } du$

where $u\quad =\quad g\left( x \right) ,\quad du\quad =\quad g^{ ' }\left( x \right) dx$

We often get an integral which does not correspond to any standard result mentioned earlier. We therefore choose a suitable new variable u to replace x.

We also replace dx by du.

If this substitution transforms the original integrand into a simpler integral in terms of u, then we integrate it with u. Finally we replace u by the corresponding value of x.

The Substitution Rule is defined as:

$\int { f\left( g\left( x \right) \right) } g^{ ' }\left( x \right) dx\quad =\quad \int { f\left( u \right) } du$

where $u\quad =\quad g\left( x \right) ,\quad du\quad =\quad g^{ ' }\left( x \right) dx$

In the case of indefinite integrals, when we use this method, we first substitute $u\quad =\quad g\left( x \right)$ , we then integrate with respect to u and then in the end we again reverse the substitution by replacing u with x and adding a constant in the end.

It is slightly more complex when there are limits involved in the case of definite integrals. We now have to change the limits of integration as well when we make the substitution, we integrate in terms of u and evaluate using the new limits of integration. This is defined as:

$\int { f\left( g\left( x \right) \right) } g^{ ' }\left( x \right) dx\quad =\quad \int { f\left( u \right) } du$

The following examples will explain how the above methods work.

Example #1

Q. Evaluate $\int { \frac { x }{ \sqrt { 2x\quad -\quad 1 } } } dx$ using the substitution $2x\quad -\quad 1\quad =\quad u$

Solution:

Let: $u\quad =\quad 2x\quad -\quad 1$

Then:

$\frac { du }{ dx } \quad =\quad 2\quad \quad \Rightarrow \quad \quad du\quad =\quad 2dx$

$dx\quad =\quad \frac { du }{ 2 }$

Substituting these values:

$\int { \frac { (u\quad +\quad 1) }{ 2\sqrt { u } } \quad \frac { du }{ 2 } }$

$=\quad \frac { 1 }{ 4 } \int { \frac { u\quad +\quad 1 }{ \sqrt { u } } \quad du }$

$=\quad \frac { 1 }{ 4 } \int { (\sqrt { u } \quad +\quad \frac { 1 }{ \sqrt { u } } )\quad du }$

$=\quad \frac { 1 }{ 4 } \int { ({ u }^{ \frac { 1 }{ 2 } }\quad +\quad { u }^{ -\frac { 1 }{ 2 } })\quad du }$

Integrating:

$=\quad \frac { 1 }{ 4 } \int { (\frac { { u }^{ { 3 }/{ 2 } } }{ \frac { 3 }{ 2 } } \quad +\quad \frac { { u }^{ { 1 }/{ 2 } } }{ \frac { 1 }{ 2 } } )\quad +\quad C }$

Replacing u by x:

$=\quad \frac { 1 }{ 4 } \left\{ \frac { 2 }{ 3 } { (2x\quad -\quad 1) }^{ \frac { 3 }{ 2 } }\quad +\quad 2{ (2x\quad -\quad 1) }^{ \frac { 1 }{ 2 } } \right\} \quad +\quad C$

Ans: $=\quad \frac { 1 }{ 6 } { (2x\quad -\quad 1) }^{ \frac { 3 }{ 2 } }\quad +\quad \frac { 1 }{ 4 } { (2x\quad -\quad 1) }^{ \frac { 1 }{ 2 } }\quad +\quad C$