4. Integration by Parts

Looking at the table of Derivative and Integral Rules, we are missing an integral version of the Product Rule. It is called Integration by Parts which is the topic of this chapter.

a1. Indefinite Integrals

In general, a derivative rule produces an indefinite integral rule: ddx[F(x)]=f(x)becomesf(x)dx=F(x)+C \dfrac{d}{dx}\left[F(x)\right]=f(x) \qquad \text{becomes} \qquad \int f(x)\,dx=F(x)+C So the Product Rule ddx[u(x)v(x)]=u(x)dvdx+v(x)dudx \dfrac{d}{dx}[u(x)v(x)]=u(x)\dfrac{dv}{dx}+v(x)\dfrac{du}{dx} produces the indefinite integral rule [u(x)dvdx+v(x)dudx]dx=u(x)v(x)+C \int \left[u(x)\dfrac{dv}{dx}+v(x)\dfrac{du}{dx}\right]\,dx=u(x)v(x)+C This integral rule is not yet in a useful form. A useful form is produced if we split up the integral on the left as the sum of two integrals and take one of them to the other side of the equation:

Integration by Parts
u(x)dvdxdx=u(x)v(x)v(x)dudxdx \int u(x)\dfrac{dv}{dx}\,dx=u(x)v(x)-\int v(x)\dfrac{du}{dx}\,dx

Notice that we have dropped the constant of integration since there is an indefinite integral on both sides which (when evaluated) will provide the constant.

This form is more useful because it helps us compute the integral of a product. If the integrand is the product of u(x)u(x) and dvdx\dfrac{dv}{dx}, then we can rewrite the integral with the integrand being the product of v(x)v(x) and dudx\dfrac{du}{dx}. Hopefully, the integral v(x)dudxdx\displaystyle \int v(x)\dfrac{du}{dx}\,dx will be easier to compute.

Frequently, students incorrectly assume:

Wrong!   The integral of a product        
      is the product of the integrals.   Wrong!

It is NOT!

By introducing the differentials du=dudxdxanddv=dvdxdx du=\dfrac{du}{dx}\,dx \qquad \text{and}\qquad dv=\dfrac{dv}{dx}\,dx the integration by parts formula may be written in a form which is easier to remember:

Integration by Parts
udv=uvvdu \int u\,dv=u\,v-\int v\,du where du=dudxdxdu=\dfrac{du}{dx}\,dx and dv=dvdxdxdv=\dfrac{dv}{dx}\,dx.

Memorize this!

Notice that the integration by parts formula does not compute the integral but rather transforms it into another integral which is hopefully easier to compute. This should be clarified by the following example:

Compute xcosxdx.\displaystyle \int x\cos x\,dx.

To apply the formula, we need to identify uu and dv.dv. If we take u=xu=x and dv=cosxdxdv=\cos x\,dx, then du=dxdu=dx and v=sinx.v=\sin x.

u=xdudx=1du=dudxdx=1dx=dxu=x \qquad \Longrightarrow \qquad \dfrac{du}{dx}=1 \qquad \Longrightarrow \qquad du=\dfrac{du}{dx}\,dx=1\,dx=dx dv=cosxdx=dvdxdxdvdx=cosxv=sinxdv=\cos x\,dx=\dfrac{dv}{dx}\,dx \qquad \Longrightarrow \qquad \dfrac{dv}{dx} =\cos x \qquad \Longrightarrow \qquad v=\sin x

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So: xcosxdx=uvvdu=xsinxsinxdx\int x\cos x\,dx=u\,v-\int v\,du=x\sin x-\int \sin x\,dx Now the final integral is easy to perform: xcosxdx=xsinx(cosx)+C=xsinx+cosx+C. \int x\cos x\,dx=x\sin x-(-\cos x)+C=x\sin x+\cos x+C .

We check by differentiating (using the Product Rule). If f(x)=xsinx+cosxf(x)=x\sin x+\cos x, then: f(x)=(sinx+xcosx)sinx=xcosx f'(x)=(\sin x+x\cos x)-\sin x=x\cos x which is the integrand we started with.

We usually write uu, dvdv, dudu and vv in an array such as: u=xdv=cosxdxdu=dxv=sinx\begin{array}{ll} u=x & dv=\cos x\,dx \\ du=dx \quad & v=\sin x \end{array} Get in this habit.

Compute the integral xsinxdx\displaystyle \int x\sin x\,dx.

Answer

xsinxdx=xcosx+sinx+C\displaystyle \int x\sin x\,dx\,=-x\cos x+\sin x+C

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Solution

Let u=xdv=sinxdxdu=dxv=cosx\begin{array}{ll} u=x & dv=\sin x\,dx \\ du=dx \quad & v=-\cos x \end{array} Then xsinxdx=xcosxcosxdx=xcosx+cosxdx=xcosx+sinx+C\begin{aligned} \int x\sin x\,dx &=-x\cos x-\int -\cos x\,dx \\ &=-x\cos x+\int \cos x\,dx \\ &=-x\cos x+\sin x+C \end{aligned}

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Check

We check (using the Product Rule). If f(x)=xcosx+sinxf(x)=-x\cos x+\sin x, then: f(x)=cosx+xsinx+cosx=xsinx f'(x)=-\cos x+x\sin x+\cos x=x\sin x which is the integrand we started with.

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Compute the integral exxdx\displaystyle \int e^x x\,dx.

Hint

The main difficulty is deciding what to take for uu and what to take for dvdv. Suppose we first try u=exdv=xdxdu=exdxv=x22\begin{array}{ll} u=e^x & dv=x\,dx \\ du=e^x\,dx \quad & v=\dfrac{x^2}{2} \end{array} Then: exxdx=exx22exx22dx \int e^x x\,dx=e^x\dfrac{x^2 }{2}-\int e^x\dfrac{x^2 }{2}\,dx This is correct but notice that the remaining integral is more complicated than the original integral. So it is better to choose the parts according to: u=xdv=exdxdu=dxv=ex\begin{array}{ll} u=x & dv=e^x\,dx \\ du=dx \quad & v=e^x \end{array}

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Answer

exxdx=xexex+C\displaystyle \int e^x x\,dx=xe^x-e^x+C

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Solution

Let: u=xdv=exdxdu=dxv=ex\begin{array}{ll} u=x & dv=e^x\,dx \\ du=dx \quad & v=e^x \end{array} Then: exxdx=xexexdx=xexex+C\begin{aligned} \int e^x x\,dx &=xe^x-\int e^x\,dx \\ &=xe^x-e^x+C \end{aligned}

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Check

Using the Product Rule, if f(x)=xexexf(x)=xe^x-e^x, then f(x)=(ex+xex)ex=xex f'(x)=(e^x+xe^x)-e^x=xe^x as desired.

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