Indefinite Integrals and u-Substitution

Indefinite Integrals and u-Substitution

Find the indefinite integral.

Example 1: $\int (3x-4)^5 dx$

• Let $u = 3x - 4$
• Then $\frac{du}{dx} = 3$
• So, $dx = \frac{du}{3}$
• Substitute:
  $\int u^5 \frac{du}{3} = \frac{1}{3} \int u^5 du$
• Integrate:
  $\frac{1}{3} \cdot \frac{u^6}{6} + C = \frac{u^6}{18} + C$
• Substitute back:
  $\frac{(3x-4)^6}{18} + C$

Example 2: $\int 6x^2 (x^3 + 4)^5 dx$

• Let $u = x^3 + 4$
• Then $\frac{du}{dx} = 3x^2$
• So, $dx = \frac{du}{3x^2}$
• Substitute:
  $\int 6x^2 u^5 \frac{du}{3x^2} = \int 2 u^5 du$
• Integrate:
  $2 \cdot \frac{u^6}{6} + C = \frac{1}{3} u^6 + C$
• Substitute back:
  $\frac{1}{3} (x^3 + 4)^6 + C$

Example 3: $\int \sqrt{\sqrt{x} - 1} dx$

• Let $u = \sqrt{x} - 1$
• Then $\frac{du}{dx} = \frac{1}{2\sqrt{x}}$
• So, $dx = 2\sqrt{x} du$
• Since $u = \sqrt{x} - 1$, then $\sqrt{x} = u + 1$, and $dx = 2(u+1) du$
• Substitute:
  $\int \sqrt{u} \cdot 2(u+1) du = 2 \int (u+1)u^{1/2} du = 2 \int (u^{3/2} + u^{1/2}) du$
• Integrate:
  $2 \left[ \frac{u^{5/2}}{5/2} + \frac{u^{3/2}}{3/2} \right] + C = 2 \left[ \frac{2}{5} u^{5/2} + \frac{2}{3} u^{3/2} \right] + C = \frac{4}{5} u^{5/2} + \frac{4}{3} u^{3/2} + C$
• Substitute back:
  $\frac{4}{5} (\sqrt{x} - 1)^{5/2} + \frac{4}{3} (\sqrt{x} - 1)^{3/2} + C$

Example 4: $\int \sin(x) e^{\cos(x)} dx$

• Let $u = \cos(x)$
• Then $\frac{du}{dx} = -\sin(x)$
• So, $dx = \frac{du}{-\sin(x)}$
• Substitute:
  $\int \sin(x) e^u \frac{du}{-\sin(x)} = - \int e^u du$
• Integrate:
  $-e^u + C$
• Substitute back:
  $-e^{\cos(x)} + C$

Example 5: $\int \cot(3x) dx$

• Let $u = 3x$
• Then $\frac{du}{dx} = 3$
• So, $dx = \frac{du}{3}$
• Substitute:
  $\int \cot(u) \frac{du}{3} = \frac{1}{3} \int \cot(u) du = \frac{1}{3} \int \frac{\cos(u)}{\sin(u)} du$
• Let $w = \sin(u)$
• Then $\frac{dw}{du} = \cos(u)$
• So, $du = \frac{dw}{\cos(u)}$
• Substitute:
  $\frac{1}{3} \int \frac{\cos(u)}{w} \frac{dw}{\cos(u)} = \frac{1}{3} \int \frac{1}{w} dw$
• Integrate:
  $\frac{1}{3} \ln |w| + C$
• Substitute back:
  $\frac{1}{3} \ln |\sin(u)| + C = \frac{1}{3} \ln |\sin(3x)| + C$

Some Tricky Examples

U-sub is used, but you must solve for $x$.

Example 6: $\int \sqrt{x+1} dx$

• Let $u = x+1$
• Then $x = u-1$ and $\frac{du}{dx} = 1$
• So, $dx = du$
• Substitute:
  $\int \sqrt{u} du = \int u^{1/2} du$
• Integrate:
  $\frac{u^{3/2}}{3/2} + C = \frac{2}{3} u^{3/2} + C$
• Substitute back:
  $\frac{2}{3} (x+1)^{3/2} + C$
  $\int \frac{x}{\sqrt{x+1}} dx = \int \frac{u-1}{\sqrt{u}} du = \int (u^{1/2} - u^{-1/2}) du$
  $= \frac{2}{3}u^{3/2} - 2u^{1/2} + C = \frac{2}{3}(x+1)^{3/2} - 2(x+1)^{1/2} + C$

Inverse Trig Can Be Confused With u-Sub

Example 7: $\int \frac{1}{\sqrt{1-4x^2}} dx$

• $\int \frac{1}{\sqrt{1-(2x)^2}} dx$
• Let $u = 2x$
• Then $\frac{du}{dx} = 2$
• So, $dx = \frac{du}{2}$
• Substitute:
  $\int \frac{1}{\sqrt{1-u^2}} \frac{du}{2} = \frac{1}{2} \int \frac{1}{\sqrt{1-u^2}} du$
• Integrate:
  $\frac{1}{2} \arcsin(u) + C$
• Substitute back:
  $\frac{1}{2} \arcsin(2x) + C$