Integration by parts (HL)

CaN i UsE dI mEThoD???

Contents

Overview

Integration by parts is the integral equivalent to the product rule. It says

∫u dv=uv−∫v du\int u\d v = uv - \int v\d u

where uu, vv are functions of xx.

It is useful when the given ∫u dv\displaystyle \int u\d v is difficult to evaluate, but ∫v du\displaystyle \int v\d u is easier.

Example: Evaluate ∫1exln⁡x dx\displaystyle\int_1^{\e} x \ln x \d x.

u=ln⁡xu = \ln x,  du=1x dx\displaystyle \d u = \frac 1x\d x

v=12x2\displaystyle v = \frac12x^2,  dv=x dx\d v = x \d x

I=(ln⁡x)12x2∣1e−∫1e12x21x dx=[12x2ln⁡x−14x2]1e=12e2−0−14e2+14=14e2+14■\begin{align*} I &= (\ln x)\frac12x^2\Big|_1^{\e} - \int_1^{\e} \frac12x^2 \frac1x \d x \\ &= \left[\frac12x^2\ln x - \frac14x^2\right]_1^{\e} \\ &= \frac12\e^2 - 0 - \frac14\e^2 + \frac14 \\ &= \frac14\e^2 + \frac14 \qed\end{align*}

There was no need to change the bounds, as long as we integrate with respect to xx.

However if you need to use substitution along with by parts, then yes change limits of integration accordingly.

Repeated use of integration by parts

The tabular method to repeated integration by parts is popularized as the “DI Method” by @blackpenredpen  (yayyyy).

Let’s look at a (modified version of a) May 2011 question and see how the tabular method is viable.

Evaluate

∫0πe−xsin⁡x dx\int_0^\pi \e^{-x} \sin x\d x

It does not really matter which one to differentiate (uu) or integrate (vv), as both repeats. Differentiate the D column; integrate the I column. The sign alternates, because of the negative sign in integration by parts.

signDI
++e−x\color{blue}\e^{-x}sin⁡x\sin x
−\color{darkgray}-−e−x\color{green}-\e^{-x}−cos⁡x\color{magenta}-\cos x
+\color{gray}+e−x\color{teal}\e^{-x}−sin⁡x\color{orange}-\sin x

The key thing is to remember is that diagonal down is uvuv, and each row is v duv \d u. Let II be the indefinite integral. Associate the sign with D.

The first step is

I=+e−x(−cos⁡x)−∫(−e−x)(−cos⁡x) dxI = +{\color{blue}\e^{-x}}({\color{magenta}-\cos x}) {\color{darkgray}-} \int {\color{green}\left(-\e^{-x}\right)}({\color{magenta}-\cos x})\d x

This gets you a mark. Then write the new diagonal down, as well as the across.

I=+e−x(−cos⁡x)−(−e−x)(−sin⁡x)+∫e−x(−sin⁡x) dxI = +{\color{blue}\e^{-x}}({\color{magenta}-\cos x}) {\color{darkgray}-} ({\color{green}-\e^{-x}})({\color{orange}-\sin x}) {\color{gray}+} \int {\color{teal}\e^{-x}}({\color{orange}-\sin x})\d x

This gets you another mark. We have found an integral that’s same as the original question (and off by a minus sign), so we stop. Collect the like terms and simplify. Technically when there is only a single indefinite integral in an equation we need the +C+C. IB generally tolerates missing intermediate +C+C unless it changes the algebra, such as when solving a differential equation.

2∫e−xsin⁡x dx=−e−xcos⁡x−e−xsin⁡x+C∫0πe−xsin⁡x dx=−12e−x(cos⁡x+sin⁡x)∣0π=−12(e−π(−1+0)−e0(1−0))=−12(−e−π−1)=12(e−π+1)■\begin{align*}2\int \e^{-x} \sin x\d x &= -\e^{-x}\cos x - \e^{-x}\sin x + C \\ \int_0^\pi \e^{-x} \sin x\d x &= -\frac12 \e^{-x}(\cos x + \sin x)\Big|_0^\pi \\ &= -\frac12 \left(\e^{-\pi}(-1 + 0) - \e^{0}(1 - 0)\right) \\ &= -\frac12 \left(- \e^{-\pi} - 1\right) \\ &= \frac12 \left(\e^{-\pi} + 1\right) \qed \end{align*}

The verdict is that as long as you correctly write the results after each use of integration by parts, you will get marks. You just cannot directly go to the answer.

Strategies

  1. If everything looks tough to integrate, use that as the vv and use  du=1 dx,u=x\d u = 1 \d x, u = x.

  2. Tabular (DI) does not necessarily work for repeated integration if uu and vv change. See the integration of

    ∫(ln⁡x)2 dx=x(ln⁡x)2−∫2ln⁡x dx=x(ln⁡x)2−2(xln⁡x−∫1 dx)=x(ln⁡x)2−2(xln⁡x−x)+C=x(ln⁡x)2−2xln⁡x+2x+C\begin{align*} \int (\ln x)^2 \d x &= x (\ln x)^2 - \int 2\ln x \d x \\ &= x (\ln x)^2 - 2\left(x\ln x - \int 1 \d x\right) \\ &= x (\ln x)^2 - 2\left(x\ln x - x\right) + C \\ &= x (\ln x)^2 - 2x\ln x + 2x + C \end{align*}

  3. Factor out constant factors (coefficients) and negative signs before integrating.