6.11 Integrating Using Integration by Parts

• Integration by Parts is a technique used to integrate products of functions
• Steps to apply Integration by Parts

  1️⃣ Choose $u$ and $dv$

  2️⃣ Find $du$ and $v$

  3️⃣ Apply the Formula

  4️⃣ Evaluate the new integral
• The Integration by Parts formula transforms complex integrals into simpler ones.
• What is the Integration by Parts formula?

  $\int u \, dv =$$uv - \int v \, du$
• In the Integration by Parts formula, du</latex> is the derivative of $u$ with respect to x
• Why is choosing $u$ and $dv$ correctly important in Integration by Parts?


  To simplify the new integral
• The new integral in Integration by Parts is always more complex than the original.

  False
• The final step in Integration by Parts is to evaluate the new integral
• What is the formula for Integration by Parts?

  $\int u \, dv =$$uv - \int v \, du$
• Integration by Parts is based on the product rule of differentiation.
• In Integration by Parts, $dv$ is typically chosen so that it is easily integrable
• What is the Integration by Parts formula?

  $\int u \, dv =$$uv - \int v \, du$
• What is the Integration by Parts formula?

  $\int u \, dv =$$uv - \int v \, du$
• By choosing $u$ and $dv$ appropriately, the new integral $\int v \, du$ is often easier to solve than the original
• Steps to solve an integral using Integration by Parts

  1️⃣ Choose $u$ and $dv$

  2️⃣ Find $du$ and $v$

  3️⃣ Apply the formula

  4️⃣ Evaluate the new integral
• dv in the Integration by Parts formula should be easily integrable.
• To find $du$, you need to differentiate $u$, and to find $v$, you need to integrate <dv</latex><dv</latex>
• What is the result of choosing $u =$$x$ and $dv =$$\cos(x) \, dx$ in $\int x \cos(x) \, dx$?


  $- x\cos(x) +$$\sin(x) +$$C$
• Choosing $u$ as the function that simplifies upon differentiation is a good strategy.
• Match the function with its derivative or integral:

  u = x ↔️ du = dx

  dv = e^{x} dx ↔️ v = e^{x}
• What is the value of $\int e^{x} dx$?


  e^{x} + C</latex>
• Integration by Parts is used to simplify integrals of products of functions.
• Steps to solve an integral using Integration by Parts

  1️⃣ Choose $u$ and $dv$

  2️⃣ Find $du$ and $v$

  3️⃣ Apply the formula

  4️⃣ Evaluate the new integral
• What is the Integration by Parts formula?

  $\int u \, dv =$$uv - \int v \, du$
• The new integral $\int v \, du$ in Integration by Parts is often easier to solve than the original.
• What is the Integration by Parts formula?

  $\int u \, dv =$$uv - \int v \, du$
• In the Integration by Parts formula, $u$ and $v$ are functions of x
• The term $du$ in the Integration by Parts formula represents the derivative of $u$ with respect to $x$.
• What does $dv$ represent in the Integration by Parts formula?


  The differential of $v$
• Steps for applying Integration by Parts

  1️⃣ Choose $u$ and $dv$

  2️⃣ Find $du$ and $v$

  3️⃣ Apply the formula

  4️⃣ Evaluate the new integral
• For the integral $\int x e^{x} dx$, we choose $u =$$x$ and dv = e^{x} dx</latex>, which simplifies to dx
• What is the derivative of $u =$$x$ in the example $\int x e^{x} dx$?


  $du =$$dx$
• What is the integral of dv = e^{x} dx</latex> in the example $\int x e^{x} dx$?


  $v =$$e^{x}$
• The final result of $\int x e^{x} dx$ is $x e^{x} - e^{x} +$$C$.
• What does the acronym LIATE stand for in the context of Integration by Parts?

  Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential
• When using LIATE, the function that appears earlier in the list is chosen as u
• The acronym LIATE helps in choosing $dv$ as the function that is easier to differentiate.


  False
• In the integral $\int x \cos(x) \, dx$, which function should be chosen as $u$?


  $u =$$x$
• How is $du$ found in the Integration by Parts process?


  By differentiating $u$
• When finding $v$, no constant of integration is needed because it cancels out in the final formula