Tutorials with examples and detailed solutions and exercises with answers on how to use the powerful technique of integration by substitution to find integrals.
Solution to Example 1:
Let \( u = ax + b \) which gives \( \dfrac{du}{dx} = a \) or \( dx = \dfrac{1}{a} du \). The substitution helps in computing the integral as follows
\( \displaystyle \int \sin(ax + b) \, dx\)
\( = \dfrac{1}{a} \int \sin(u) \, du \)
\( = -\dfrac{1}{a} \cos(u) + C \)
\( = -\dfrac{1}{a} \cos(ax + b) + C\)
Solution to Example 2:
Let \( u = 3x - 2 \) which gives \( \dfrac{du}{dx} = 3 \) or \( dx = \dfrac{1}{3} du \). Hence
\( \displaystyle \int e^{3x - 2} \, dx\)
\( = \displaystyle \int e^{u} \cdot \dfrac{1}{3} \, du \)
\( = \dfrac{1}{3} e^{u} \)
\( = \dfrac{1}{3} e^{3x - 2} + C\)
Solution to Example 3:
Let \( u = 2x^2 + 5 \) which gives \( \dfrac{du}{dx} = 4x \), \( du = 4x \, dx \), \( \dfrac{1}{4} du = x \, dx \). The substitution gives
\( \displaystyle \int x (2x^2 + 5)^4 \, dx\)
\( = \displaystyle \int \dfrac{1}{4} u^4 \, du \)
\( = \dfrac{1}{20} u^5 \)
\( = \dfrac{1}{20} (2x^2 + 5)^5 + C \)
Solution to Example 4:
Let \( u = 2x + 1 \) which gives \( \dfrac{du}{dx} = 2 \) and \( dx = \dfrac{1}{2} du \). Solve \( u = 2x + 1 \) for \( x \) to obtain \( x = \dfrac{1}{2}(u - 1) \). The substitution gives
\( \displaystyle \int x \sqrt{2x + 1} \, dx\)
\( = \displaystyle \int \dfrac{1}{2}(u - 1) \sqrt{u} \cdot \dfrac{1}{2} \, du \)
\( = \dfrac{1}{4} \int (u - 1) u^{1/2} \, du \)
\( = \dfrac{1}{4} \left( \dfrac{2}{5} u^{5/2} - \dfrac{2}{3} u^{3/2} \right) \)
\( = \dfrac{(2x + 1)^{3/2}(3x - 1)}{15} + C \)
Solution to Example 5:
Let \( u = x - 5 \) which gives \( \dfrac{du}{dx} = 1 \). Substituting into the given integral, we obtain
\( \displaystyle \int (x - 5)^{-4} \, dx\)
\( = \displaystyle \int u^{-4} \, du \)
\( = -\dfrac{1}{3} u^{-3} \)
\( = -\dfrac{1}{3}(x - 5)^{-3} + C \)
Solution to Example 6:
Let \( u = x^2 + 2 \) which gives \( \dfrac{du}{dx} = 2x \) and \( \dfrac{1}{2} du = x \, dx \). A substitution into the given the integral gives
\( \displaystyle \int -x e^{x^2 + 2} \, dx\)
\( = \displaystyle \int - e^{u} \cdot \dfrac{1}{2} \, du \)
\( = -\dfrac{1}{2} \int e^{u} \, du \)
\( = -\dfrac{1}{2} e^{u} \)
\( = -\dfrac{1}{2} e^{x^2 + 2} + C \)
Solution to Example 7:
Let \( u = \sin(x) \) which gives \( \dfrac{du}{dx} = \cos(x) \) or \( \cos(x) \, dx = du \). Substitute into the integral to obtain
\( \displaystyle \int \cos(x) \sin^4(x) \, dx\)
\( = \displaystyle \int u^4 \, du \)
\( = \dfrac{1}{5} u^5 \)
\( = \dfrac{1}{5} \sin^5(x) + C \)
Solution to Example 8:
Let \( u = 4x + 1 \) which gives \( \dfrac{du}{dx} = 4 \) or \( dx = \dfrac{1}{4} du \). Solve \( u = 4x + 1 \) for \( x \) to obtain \( x = \dfrac{1}{4}(u - 1) \). Substitute to obtain
\( \displaystyle \int \dfrac{3x}{4x + 1} \, dx\)
\( = \displaystyle \int 3 \cdot \dfrac{1}{4} \cdot \dfrac{u - 1}{u} \, du \)
\( = \dfrac{3}{16} \displaystyle \int \dfrac{u - 1}{u} \, du \)
\( = \dfrac{3}{16} \displaystyle \int \left( 1 - \dfrac{1}{u} \right) \, du \)
\( = \dfrac{3}{16}(u - \ln|u|) \)
\( = \dfrac{3}{16}(4x + 1 - \ln|4x + 1|) + C \)
Solution to Example 9:
Let \( u = x - 2 \) which gives \( \dfrac{du}{dx} = 1 \), \( dx = du \) and \( x = u + 2 \). Substitution
\( \displaystyle \int \dfrac{x}{\sqrt{x - 2}} \, dx\)
\( = \displaystyle \int \dfrac{u+2}{\sqrt{u}} \, dx \)
\( = \int (u^{1/2} + 2u^{-1/2}) \, dx \)
\( = \dfrac{2}{3}u^{3/2} + 4 \sqrt{u} \)
\( = \dfrac{2}{3}(x - 2)^{3/2} + 4 \sqrt{x - 2} + C \)
Solution to Example 10:
Let \( u = x + 2 \) which gives \( \dfrac{du}{dx} = 1 \), \( dx = du \) and also \( x = u - 2 \). Using the above substitution we obtain
\( \displaystyle \int (x + 2)^3(x + 4)^2 \, dx\)
\( = \displaystyle \int u^3(u + 4)^2 \, du \)
\( = \displaystyle \int (u^5 + 4u^4 + 4u^3) \, du \)
\( = \dfrac{1}{6} u^6 + \dfrac{4}{5}u^5 + u^4 \)
\( = \dfrac{1}{6}(x + 2)^6 + \dfrac{4}{5}(x + 2)^5 + (x + 2)^4 + C \)
Solution to Example 11:
Let \( u = x^2 + 3x + 1 \) which gives \( \dfrac{du}{dx} = 2x + 3 \) or \( (2x + 3) \, dx = du \). The substitution helps in computing the integral as follows
\(\displaystyle \int \dfrac{1}{u} \, du\)
\( = \ln|u| \)
\( = \ln |x^2 + 3x + 1| + C \)