Goal: Not all integrals can be evaluated directly using standard antiderivative rules like power, exponential, or trigonometric formulas. The substitution rule helps rewrite integrals into a form where these rules apply.
1. Indefinite Integrals
Formula:
\[
\int f(g(x))g'(x)\,dx = \int f(u)\,du, \quad u = g(x)
\]
Example 1
\[I=
\int 2x\sqrt{1+x^2}\,dx
\]
Solution:
Let \(u = 1 + x^2\), so \(du = 2x\,dx\).
\[I
= \int \sqrt{u}\,du = \frac{2}{3}u^{3/2} + C
\]
\[
= \frac{2}{3}(1+x^2)^{3/2} + C
\]
Example 2
\[I=
\int x^3 \cos(x^4 + 2)\,dx
\]
Solution:
Let \(u = x^4 + 2\), so \(du = 4x^3 dx\).
\[I
= \frac{1}{4} \int \cos(u)\,du = \frac{1}{4}\sin(u) + C
\]
\[
= \frac{1}{4}\sin(x^4 + 2) + C
\]
Example 3
\[I=
\int \sqrt{2x+1}\,dx
\]
Solution:
Let \(u = 2x+1\), so \(du = 2dx\).
\[I
= \frac{1}{2}\int u^{1/2}du = \frac{1}{3}u^{3/2} + C
\]
\[
= \frac{1}{3}(2x+1)^{3/2} + C
\]
Example 4
\[I=
\int x^5 \sqrt{1+x^2}\,dx
\]
Solution:
Let \(u = 1+x^2\), so \(du = 2x dx\).
Rewrite \(x^5 = x^4 \cdot x\) and use \(x^2 = u-1\).
\[I
= \frac{1}{2} \int (u-1)^2 u^{1/2} du
\]
\[
= \frac{1}{2} \int (u^{5/2} - 2u^{3/2} + u^{1/2}) du
\]
\[
= \frac{1}{7}u^{7/2} - \frac{2}{5}u^{5/2} + \frac{1}{3}u^{3/2} + C
\]
\[
= \frac{1}{7}(1+x^2)^{7/2}
- \frac{2}{5}(1+x^2)^{5/2}
+ \frac{1}{3}(1+x^2)^{3/2} + C
\]
2. Definite Integrals
\[
\int_a^b f(g(x))g'(x)\,dx
=
\int_{g(a)}^{g(b)} f(u)\,du
\]
Example 1
\[I=
\int_0^1 \sqrt{2x+1}\,dx
\]
Solution:
Let \(u = 2x+1\).
Change limits: \(x=0 \to u=1\), \(x=1 \to u=3\).
\[I
= \frac{1}{2} \int_1^3 u^{1/2}du
= \frac{1}{3}(3^{3/2} - 1)
\]
Example 2
\[I=
\int_1^e \frac{\ln x}{x}\,dx
\]
Solution:
Let \(u = \ln x\), so \(du = \frac{1}{x}dx\).
Change limits: \(x=1 \to u=0\), \(x=e \to u=1\).
\[I
= \int_0^1 u\,du = \frac{1}{2}
\]