17.3 Integration by Parts

17.17   Theorem (Integration by parts.) Let $f,g$ be functions that are continuous on an interval $[a,b]$ and differentiable on $(a,b)$. If $f'g$ has an antiderivative on $[a,b]$, then $g'f$ also has an antiderivative on $[a,b]$ and

$$\int g'f = fg - \int f'g.$$ (17.18)

We call formula (17.18) the formula for integration by parts.

Proof:     This theorem is just a restatement of the product rule for differentiation. If $f$ and $g$ are differentiable on $(a,b)$ then the product rule says that

$$(fg)' = f'g + g'f$$

so that

$$g'f = (fg)' - f'g$$

on $(a,b)$. If $\int f'g$ is an antiderivative for $f'g$ on $[a,b]$, then $fg - \int f'g$ is continuous on $[a,b]$, and

$$(fg - \int f'g)' = (fg)' - f'g = g'f$$

on $(a,b)$. We have shown that $fg - \int f'g$ is an antiderivative for $g'f$ on $[a,b]$. $\diamondsuit$

17.19   Example. We will calculate $${\int_0^{\pi} x \sin(3x)dx}$$. We begin by searching for an antiderivative for $x \sin(3x)$. Let

$$f(x) = x,\mbox{{}}$$

$$g'(x) = \sin(3x), \mbox{{}}$$

$$f'(x) = 1,\mbox{{}}$$

$$g(x) = -\frac{1}{3}\cos(3x).\mbox{{}}$$

Then by the formula for integration by parts

$$\int x \sin(3x) dx = \int f(x)g'(x) dx \mbox{{}}$$

$$= f(x)g(x) - \int f'(x) g(x) dx \mbox{{}}$$

$$= -\frac{x}{3}\cos(3x) + \frac{1}{3} \int \cos(3x) dx \mbox{{}}$$

$$= -\frac{x}{3} \cos(3x) + \frac{1}{9} \sin(3x).$$ (17.20)

Hence

$$\int _0 ^{\pi} x \sin(3x) dx = (\left. -\frac{x}{3} \cos(3x) + \frac{1}{9} \sin(3x) )\right\vert _0^{\pi} \mbox{{}}$$

$$= -\frac{\pi}{3}\cos(3\pi) = \frac{\pi}{3}. \mbox{{}}$$

Suppose I had tried to find $\int x\sin(3x)$ in the following way: Let

$$f(x) = \sin(3x), \mbox{{}}$$

$$g'(x) = x,\mbox{{}}$$

$$f'(x) = 3\cos(3x), \mbox{{}}$$

$$g(x) = \frac{1}{2}x^2. \mbox{{}}$$

Then by the formula for integration by parts

$$\int x \sin(3x) dx = \int f(x)g'(x) dx \mbox{{}}$$

$$= f(x)g(x) - \int f'(x) g(x) dx \mbox{{}}$$

$$= \frac{1}{2}x^2 \sin(3x) - \frac{3}{2} \int x^2 \cos(3x) dx.$$ (17.21)

In this case the antiderivative $\int x^2 \cos(3x) dx$ looks more complicated than the one I started out with. When you use integration by parts, it is not always clear what you should take for $f$ and for $g'$. If you find things are starting to look more complicated rather than less complicated, you might try another choice for $f$ and $g'$.

Integration by parts is used to evaluate antiderivatives of the forms
$${\int x^n\sin(ax)dx}$$, $${\int x^n\cos(ax)dx}$$, and $${\int x^n e^x dx}$$ when $n$ is a positive integer. These can be reduced to antiderivatives of the forms $${\int x^{n-1}\sin(ax)dx}$$,
$${\int x^{n-1}\cos(ax)dx}$$, and $${\int x^{n-1} e^x dx}$$, so by applying the process $n$ times we get the power of $x$ down to $x^0$, which gives us antiderivatives we can easily find.

17.22   Example. We will calculate $\int \sin(\ln(x)).$

Let

$$f(x) = \sin(\ln(x)), \mbox{{}}$$

$$g'(x) = 1,\mbox{{}}$$

$$g(x) = x,\mbox{{}}$$

$$f'(x) = \frac{\cos(\ln(x))}{x} .\mbox{{}}$$

Then

$$\int \sin(\ln(x))dx = \int f(x) g'(x) dx =f(x)g(x) -\int f'(x)g(x)dx \mbox{{}}$$

$$= x \sin(\ln(x)) - \int \cos(\ln(x))dx$$ (17.23)

We will now use integration by parts to find an antiderivative for $\cos(\ln(x))$. Let

$$F(x) = \cos(\ln(x)), \mbox{{}}$$

$$G'(x) = 1,\mbox{{}}$$

$$G(x) = x,\mbox{{}}$$

$$F'(x) = -\frac{\sin(\ln(x))}{x}.\mbox{{}}$$

Then

$$\int \cos(\ln(x))dx = \int F(x) G'(x) = F(x)G(x) - \int F'(x)G(x) dx \mbox{{}}$$

$$= x \cos(\ln(x)) + \int \sin(\ln(x))dx$$ (17.24)

From equations (17.23) and (17.24) we see that

$$\int \sin(\ln(x))dx = x \sin(\ln(x)) -\left(x\cos(\ln(x)) +\int\sin(\ln(x)) \right).$$

Thus

$$2 \int \sin(\ln(x))dx = x \sin(\ln(x)) - x\cos(\ln(x)),$$

and

$$\int \sin(\ln(x))dx = \frac{x}{2} (\sin(\ln(x)) - \cos(\ln(x))).$$

17.25   Example. We will calculate $\int \ln(t)dt$. Let

$$f(t) = \ln(t), \mbox{{}}$$

$$g'(t) = 1,\mbox{{}}$$

$$g(t) = t, \mbox{{}}$$

$$f'(t) = \frac{1}{t}. \mbox{{}}$$

Then

$$\int \ln(t) dt = \int f(t)g'(t)dt = f(t)g(t) - \int f'(t)g(t)dt \mbox{{}}$$

$$= t \ln(t) - \int 1 dt \mbox{{}}$$

$$= t \ln(t) -t. \mbox{{}}$$

17.26   Theorem (Antiderivative of inverse functions.) Let $I$ and $J$ be intervals in $\mbox{{\bf R}}$, and let $f\colon I\to J$ be a continuous function such that $f'(x)$ is defined and non-zero for all $x$ in the interior of $I$. Suppose that $g\colon J\to I$ is the inverse function for $f$, and that $F$ is an antiderivative for $f$. Then an antiderivative for $g$ on $J$ is given by

$$\int g(x)\; dx=xg(x)-(F\circ g)(x).$$ (17.27)

Proof: Let $h(x)=x$. Then $h'(x)=1$, and

$$\int g(x)\; dx = \int g(x) h'(x) dx = g(x)h(x)-\int g'(x)h(x) dx$$

$$= xg(x) -\int x g'(x) dx$$ (17.28)

Now $F'=f$ and $f\circ g(x)=x$ for all $x$ in $J$, so

$$(F\circ g)'(x)=F'\Big(g(x)\Big)\cdot g'(x)=f\Big(g(x)\Big)\cdot g'(x)=xg'(x)$$

and it follows from (17.28) that

$$\begin{eqnarray*} \int g(x)\;dx &=& xg(x)-\int(F\circ g)'(x)\; dx \\ &=& xg(x)-(F\circ g)(x).\mbox{ $\diamondsuit$} \end{eqnarray*}$$

Remark: It follows from the proof of the previous theorem that if you know an antiderivative for a function $f$, then you can find an antiderivative for the inverse function $g$ by integration by parts. This is what you should remember about the theorem. The formula (17.27) is not very memorable.

17.29   Example. In the previous theorem, if we take

$$f(x)=e^x,\;\; F(x)=e^x,\;\; g(x)=\ln (x),$$

then we get

$$\int\ln(x)\;dx=x\ln(x)-e^{\ln(x)}=x\ln(x)-x.$$

This agrees with the result obtained in example 17.25.

17.30   Exercise. What is wrong with the following argument? Let

$$f(x) = \frac{1}{x}, \mbox{{}}$$

$$g'(x) = 1,\mbox{{}}$$

$$f'(x) = -\frac{1}{x^2}, \mbox{{}}$$

$$g(x) = x.\mbox{{}}$$

Then

$$\int\frac{1}{x}dx = \int f(x) g'(x) dx =f(x)g(x) -\int f'(x)g(x)dx \mbox{{}}$$

$$= 1 + \int \frac{1}{x}dx.\mbox{{}}$$

If we subtract $${\int \frac{1}{x}dx}$$ from both sides we obtain

$$0 = 1.$$

17.31   Exercise. A Calculate the following antiderivatives:

a)

${\displaystyle \int x e^x dx.}$

b)

${\displaystyle \int e^x \sin(x) dx.}$ (Integrate by parts more than once.)

c)

${\displaystyle \int \arctan(u) du.}$

d)

${\displaystyle \int \frac{x}{\sqrt{4-x^2}} dx.}$

e)

${\displaystyle \int x \sqrt{4-x^2} dx. }$

f)

${\displaystyle \int x^r \ln(\vert x\vert) dx}$, where $r\in\mbox{{\bf R}}$. Have you considered the case where $r=-1$?

g)

${\displaystyle \int x^2 \cos(2x) dx}$.