Algebraic Substitution Integration Technique
Smaridasa Mathematics – The Algebraic Substitution Integration Technique is usually used when basic integration formulas (both algebraic and trigonometric) cannot directly solve the problem. Although named "Algebraic Substitution", this technique can also be applied to trigonometric integrals.
Concept of Algebraic Substitution
As the name suggests, algebraic substitution means we substitute a function with a specific algebraic form to make integration easier.
Suppose we have an integral of the form \( \int [f(x)]^\alpha g(x) \, dx \) that is difficult to integrate directly with basic formulas. We substitute by setting:
\[ u = f(x) \]
The derivative of \( u \) is:
\[ \frac{du}{dx} = f'(x) \quad \Rightarrow \quad dx = \frac{du}{f'(x)} \]
The integral then becomes:
\[ \int [f(x)]^\alpha g(x) \, dx = \int u^\alpha g(x) \frac{du}{f'(x)} \]
Key Point: Algebraic substitution is successful if the derivative \( f'(x) \) cancels with part of \( g(x) \), leaving an integrable expression in terms of \( u \).
Example Problems
1. \( \displaystyle \int 2x (4x^2 + 5)^{15} \, dx \)
Solution:
Let \( u = 4x^2 + 5 \), then \( du = 8x \, dx \) or \( dx = \frac{du}{8x} \).
\[ \int 2x (4x^2 + 5)^{15} \, dx = \int 2x \cdot u^{15} \cdot \frac{du}{8x} = \int \frac{1}{4} u^{15} \, du \]
\[ = \frac{1}{4} \cdot \frac{1}{16} u^{16} + c = \frac{1}{64} u^{16} + c \]
Back substitute \( u = 4x^2 + 5 \):
\[ = \frac{1}{64} (4x^2 + 5)^{16} + c \]
2. \( \displaystyle \int (4x + 8) \sqrt{x^2 + 4x - 5} \, dx \)
Solution:
Let \( u = x^2 + 4x - 5 \), then \( du = (2x + 4) \, dx = 2(x + 2) \, dx \).
Note: \( 4x + 8 = 4(x + 2) \).
\[ \int (4x + 8) \sqrt{x^2 + 4x - 5} \, dx = \int 4(x + 2) \sqrt{u} \cdot \frac{du}{2(x + 2)} \]
\[ = \int 2 \sqrt{u} \, du = 2 \cdot \frac{2}{3} u^{3/2} + c = \frac{4}{3} u^{3/2} + c \]
Back substitute \( u = x^2 + 4x - 5 \):
\[ = \frac{4}{3} (x^2 + 4x - 5)^{3/2} + c \]
3. \( \displaystyle \int \frac{3x - 1}{\sqrt{3x^2 - 2x + 7}} \, dx \)
Solution:
Let \( u = 3x^2 - 2x + 7 \), then \( du = (6x - 2) \, dx = 2(3x - 1) \, dx \).
\[ \int \frac{3x - 1}{\sqrt{3x^2 - 2x + 7}} \, dx = \int \frac{3x - 1}{\sqrt{u}} \cdot \frac{du}{2(3x - 1)} \]
\[ = \int \frac{1}{2} u^{-1/2} \, du = \frac{1}{2} \cdot 2 u^{1/2} + c = \sqrt{u} + c \]
Back substitute \( u = 3x^2 - 2x + 7 \):
\[ = \sqrt{3x^2 - 2x + 7} + c \]
4. \( \displaystyle \int \frac{5}{\sqrt{x}} \sqrt{(\sqrt{x} + 2)^3} \, dx \)
Solution:
Let \( u = \sqrt{x} + 2 \), then \( du = \frac{1}{2\sqrt{x}} \, dx \) or \( dx = 2\sqrt{x} \, du \).
\[ \int \frac{5}{\sqrt{x}} \sqrt{(\sqrt{x} + 2)^3} \, dx = \int \frac{5}{\sqrt{x}} \sqrt{u^3} \cdot 2\sqrt{x} \, du \]
\[ = \int 10 u^{3/2} \, du = 10 \cdot \frac{2}{5} u^{5/2} + c = 4 u^{5/2} + c \]
Back substitute \( u = \sqrt{x} + 2 \):
\[ = 4 (\sqrt{x} + 2)^{5/2} + c \]
5. \( \displaystyle \int 6x^2 \sin(3x^3) \, dx \)
Solution:
Let \( u = 3x^3 \), then \( du = 9x^2 \, dx \) or \( dx = \frac{du}{9x^2} \).
\[ \int 6x^2 \sin(3x^3) \, dx = \int 6x^2 \sin u \cdot \frac{du}{9x^2} = \int \frac{2}{3} \sin u \, du \]
\[ = \frac{2}{3} (-\cos u) + c = -\frac{2}{3} \cos(3x^3) + c \]
6. \( \displaystyle \int \frac{\cos(\sqrt{x} + 4)}{\sqrt{x}} \, dx \)
Solution:
Let \( u = \sqrt{x} + 4 \), then \( du = \frac{1}{2\sqrt{x}} \, dx \) or \( dx = 2\sqrt{x} \, du \).
\[ \int \frac{\cos(\sqrt{x} + 4)}{\sqrt{x}} \, dx = \int \frac{\cos u}{\sqrt{x}} \cdot 2\sqrt{x} \, du \]
\[ = \int 2 \cos u \, du = 2 \sin u + c = 2 \sin(\sqrt{x} + 4) + c \]
7. \( \displaystyle \int \frac{\sec^2(2 - \frac{1}{\sqrt{x}})}{2\sqrt{x^3}} \, dx \)
Solution:
Let \( u = 2 - \frac{1}{\sqrt{x}} \), then \( du = \frac{1}{2\sqrt{x^3}} \, dx \) or \( dx = 2\sqrt{x^3} \, du \).
\[ \int \frac{\sec^2(2 - \frac{1}{\sqrt{x}})}{2\sqrt{x^3}} \, dx = \int \frac{\sec^2 u}{2\sqrt{x^3}} \cdot 2\sqrt{x^3} \, du \]
\[ = \int \sec^2 u \, du = \tan u + c = \tan\left(2 - \frac{1}{\sqrt{x}}\right) + c \]
General Formulas via Substitution
For \( n \neq -1 \):
\[ \int k(ax + b)^n \, dx = \frac{k}{a} \cdot \frac{1}{n+1} (ax + b)^{n+1} + c \]
Example 8: \( \displaystyle \int 4(2x - 5)^{31} \, dx \)
\[ = \frac{4}{2} \cdot \frac{1}{32} (2x - 5)^{32} + c = \frac{1}{16} (2x - 5)^{32} + c \]
For \( n = -1 \):
\[ \int \frac{k}{ax + b} \, dx = \frac{k}{a} \ln|ax + b| + c \]
Example 9: \( \displaystyle \int \frac{3}{2x - 5} \, dx \)
\[ = \frac{3}{2} \ln|2x - 5| + c \]
Example 10: \( \displaystyle \int \sqrt{3x + 2} \, dx \)
Rewrite as \( \int (3x + 2)^{1/2} \, dx \) and use the formula with \( k=1, a=3, b=2, n=\frac{1}{2} \):
\[ = \frac{1}{3} \cdot \frac{1}{\frac{1}{2} + 1} (3x + 2)^{3/2} + c = \frac{2}{9} (3x + 2)^{3/2} + c \]