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# 8.2: Integration by Parts

Difficulty Level: At Grade Created by: CK-12

This activity is intended to supplement Calculus, Chapter 7, Lesson 2.

In this activity, you will explore:

• product rule of differentiation
• integration by parts

## Exercises

1. State the product rule for a function of the form u(x)v(x)\begin{align*}u(x)*v(x)\end{align*}.

2. Apply the product rule to the function sin(x)ln(x)\begin{align*}\sin(x)* \ln(x)\end{align*}.

3. Do you agree or disagree with the following statement? Explain.

ddx(f(x))dx=ddx(f(x)dx)=f(x)\begin{align*}\int\limits \frac{d}{dx} (f(x)) dx = \frac{d}{dx} \left (\int\limits f(x)dx \right ) = f(x)\end{align*}

4. What is the integral of the left side of the product rule?

(ddx(u(x)v(x))dx=\begin{align*}\int\limits \left (\frac{d}{dx} (u(x) \cdot v(x) \right )dx = \end{align*}

5. What is the integral of the right side?

(u(x)dvdx+v(x)dudx)dx=\begin{align*}\int\limits \left (u (x) \cdot \frac{dv}{dx} + v(x) \cdot \frac{du}{dx} \right )dx = \end{align*}

6. Explain the relationship between the areas shown on the graph and the following equation:

v1v2udv=uvu1u2vdu\begin{align*}\int\limits_{v_1}^{v_2} u \cdot dv = u \cdot v - \int\limits_{u_1}^{u_2} v \cdot du\end{align*}

7. Use the method of integration by parts to compute the integral of ln(x)\begin{align*}\ln (x)\end{align*}.

Remember the formula for Integration by parts is udv=uvvdu\begin{align*}\int\limits u \cdot dv = u \cdot v - \int\limits v \cdot du\end{align*}

ln(x)1 dxu=ln(x) and dv=1 dx du=v=Result=\begin{align*}& \int\limits \ln(x) \cdot 1\ dx \rightarrow u = \ln(x) \ \text{and} \ dv = 1\ dx \\ & \qquad \qquad \qquad \quad \ du = \qquad \qquad \quad v = \\ & \text{Result} = \end{align*}

Check by integration directly. (Home > F3:Calc > 2:Integrate) or (Home > 2nd\begin{align*}2^{nd}\end{align*} 7 )

Consider the function f(x)=sin(ln(x))\begin{align*}f(x) = \sin(\ln(x))\end{align*}.

udv=sin(ln(x))du=cos(ln(x))xdx=dxv=x(+C)\begin{align*}u & = \sin(\ln(x)) \rightarrow du = \frac{\cos(\ln(x))}{x} dx \\ dv & = dx \rightarrow v = x (+C)\end{align*}

sin(ln(x))1 dx=xsin(ln(x))xcos(ln(x))xdx(+C)=xsin(ln(x))cos(ln(x))dx(+C)\begin{align*}\int\limits \sin(\ln(x)) \cdot 1\ dx & = x \cdot \sin(\ln(x)) - \int\limits x \cdot \frac{\cos(\ln(x))}{x} dx (+C) \\ & = x \cdot \sin(\ln(x)) - \int\limits \cos (\ln(x))dx (+C)\end{align*}

8. Find cos(ln(x))dx\begin{align*}\int\limits \cos (\ln(x))dx\end{align*}.

ucos(ln(x))dx==du=dv=v=\begin{align*}u & = && du = && dv = && v = \\ \int\limits \cos (\ln(x))dx & =\end{align*}

9. Substitute the result for cos(ln(x))\begin{align*}\cos(\ln(x))\end{align*} into the result for sin(ln(x))\begin{align*}\sin(\ln(x))\end{align*}.

usin(ln(x))dx==du=dv=v=\begin{align*}u & = && du = && dv = && v = \\ \int\limits \sin (\ln(x))dx & =\end{align*}

10. Use integration by parts to solve the following. If you need to use integration by parts more than once, do so. Check your result.

a. tan1(x) dx\begin{align*}\int\limits \tan^{-1}(x)\ dx\end{align*}

b. x2ex dx\begin{align*}\int\limits x^2 \cdot e^x \ dx\end{align*}

c. xtan1(x) dx\begin{align*}\int\limits x \cdot \tan^{-1}(x) \ dx\end{align*}

d. xcos(2x+1) dx\begin{align*}\int\limits x \cdot \cos (2x+1)\ dx\end{align*}

11. (Extension 1) Does it matter in which order u(x)\begin{align*}u(x)\end{align*} and v(x)\begin{align*}v(x)\end{align*} are selected for the method of integration by parts?

12. (Extension 2) Is there likely to be an integration rule based upon the quotient rule just as Integration by Parts was based upon the product rule?

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