Integration by Parts

The purpose of integration by parts is to replace a difficult integral with one that is easier to evaluate. The formula that allows us to do this is $\displaystyle \int u\, dv=uv-\int v\,du.$

Suppose we are trying to do the integration $\int xe^x dx.$ We notice that $u$-substitution cannot be used, since neither $x$ nor $e^x$ is close to being the derivative of the other. A function which is the product of two different kinds of functions, like $xe^x,$ requires a new technique in order to be integrated, which is integration by parts. The rule is as follows:

$$\int u \, dv=uv-\int v \, du$$

This might look confusing at first, but it's actually very simple. Let's take a look at its proof and find out how easy and convenient it is:

Remember the product rule? It is the rule for differentiating the product of two functions, expressed in terms of

$$\big(f(x)g(x)\big)'=f'(x)g(x)+f(x)g'(x).$$

Integration by parts is exactly its antiderivative form. Integrating both sides gives

$$\begin{aligned} f(x)g(x)&=\int f'(x)g(x)\, dx+\int f(x)g'(x)\, dx\\ \Rightarrow \int f(x)g'(x)\, dx&=f(x)g(x)-\int f'(x)g(x)\, dx. \end{aligned}$$

Let $u=f(x)$ and $v=g(x).$ Then we have a more compact expression:

$$\begin{aligned} \int f(x)g'(x)\, dx&=f(x)g(x)-\int f'(x)g(x)\, dx\\ \Rightarrow \int udv &=uv-\int vdu.\ _\square \end{aligned}$$

Now that we know the rule, let's find the answer to the example above.

What is $\displaystyle{\int xe^xdx}?$

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Let $u=x$ and $v'=e^x.$ Then we have

$$\begin{aligned} u'&=1\\ v&=\int e^xdx=e^x\\ \Rightarrow \int xe^xdx&=\int u\, dv\\ &=uv-\int v\, du\\ &=xe^x-\int e^x\cdot1\, dx\\ &=xe^x-e^x+C, \end{aligned}$$

where $C$ is the constant of integration. $_\square$

As shown in the example above, we let one factor be $u$ and the other $v'$ $($or $dv).$ Then our given problem will be $\displaystyle{\int u \, dv},$ and we can apply the rule of integration by parts. So then, what is the criterion to determine which of the factors will be $u?$ Why can't we let $u=e^x$ and $v'=x?$ There is a rule to this too: Since $v'$ is to be integrated, let $v'$ be the easier to integrate. Easier means that the function changes less after integration. For instance, an exponential function, say $e^x,$ is easy to integrate, since it doesn't change at all after integration. A trigonometric function will change to its counterpart. For instance, $\sin x$ will change to $-\cos x.$ Logarithmic functions are the most difficult to integrate. Here is an explicit demonstration of the rule:

$$\begin{aligned} &&\text{Logs} &&\text{Polynomials} &&\text{Trigs} &&\text{Exponentials}\\ &&u\ \ \ \ &&\cdots\ \ \ \ \ \ &&\cdots\ \ &&v'\ \ \ \ \ \ \end{aligned}$$

The further to the right, the better to put as $v',$ and the more on the left, the better to put as $u.$ Let's see a few more examples.

Evaluate $\displaystyle\int xe^x dx.\$

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$x$ has a pretty simple derivative, so let's say $u=x$. Then $dv=e^x dx, du=dx$, and $v=\displaystyle\int dv=e^x$, which implies.

$$\displaystyle\int xe^x dx=(x)(e^x)-\displaystyle\int (e^x)(dx)=xe^x-e^x + C=e^x(x-1) + C,$$

where $C$ is the constant of integration. You can take the derivative to see that it is indeed our desired result. $_\square$

Find the indefinite integral $\displaystyle{\int x\sin x \, dx.}$

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According to the rule above, we should let $u=x$ and $v'=\sin x.$ Then we have

$$\begin{aligned} u'&=1\\ v&=-\cos x\\ \Rightarrow\int x\sin x \, dx&=\int u \, dv\\ &=uv-\int v \, du\\ &=-x\cos x-\int (-\cos x)\cdot1 \, dx\\ &=-x\cos x+\sin x+C, \end{aligned}$$

where $C$ is the constant of integration. $_\square$

Find the indefinite integral $\displaystyle{\int x^2\cos x \, dx.}$

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According to the rule above, we should let $u=x^2$ and $v'=\cos x.$ Then we have

$$\begin{aligned} u'&=2x\\ v&=\sin x\\ \Rightarrow\int x^2\cos x \, dx&=\int u \, dv\\ &=uv-\int v \, du\\ &=x^2\sin x-\int 2x\sin x\, dx. \end{aligned}$$

We know what $\displaystyle{\int x\sin x\, dx}$ is from an example question above (otherwise, we would have to use integration by parts one more time). Therefore, we have

$$x^2\sin x-\int 2x\sin x\, dx=x^2\sin x+2x\cos x-2\sin x+C,$$

where $C$ is the constant of integration.

Notice that we needed to use integration by parts twice to solve this problem. $_\square$

Find the indefinite integral $\displaystyle{\int xe^{2x} dx.}$

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According to the rule above, we should let $u=x$ and $v'=e^{2x}.$ Then we have

$$\begin{aligned} u'&=1\\ v&=\frac{1}{2}e^{2x}\\ \Rightarrow\int xe^{2x} dx&=\int u \, dv\\ &=uv-\int v \, du\\ &=\frac{1}{2}xe^{2x}-\int \frac{1}{2}e^{2x} \, dx\\ &=\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}+C, \end{aligned}$$

where $C$ is the constant of integration. $_\square$

Find the indefinite integral $\displaystyle{\int \ln x\,dx.}$

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This one needs a trick. Think of $\ln x$ as $1\cdot\ln x.$ Then, according to the rule above, we should let $u=\ln x$ and $v'=1.$ Then we have

$$\begin{aligned} u'&=\frac{1}{x}\\ v&=x\\ \Rightarrow\int \ln x \, dx&=\int u \, dv\\ &=uv-\int v \, du\\ &=x\ln x-\int x\cdot\frac{1}{x} \, dx\\ &=x\ln x-x+C, \end{aligned}$$

where $C$ is the constant of integration.

This one will appear very frequently, so memorizing $\displaystyle{\int\ln x\,dx=x\ln x-x+C}$ is highly recommended. $_\square$

Find the indefinite integral $\displaystyle{\int x\ln x\,dx.}$

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According to the rule above, we should let $u=\ln x$ and $v'=x.$ Then we have

$$\begin{aligned} u'&=\frac{1}{x}\\ v&=\frac{1}{2}x^2\\ \Rightarrow\int x\ln x \, dx&=\int u \, dv\\ &=uv-\int v \, du\\ &=\frac{1}{2}x^2\ln x-\int \frac{1}{2}x^2\cdot\frac{1}{x} \, dx\\ &=\frac{1}{2}x^2\ln x-\int\frac{1}{2}x \, dx\\ &=\frac{1}{2}x^2\ln x-\frac{1}{4}x^2+C, \end{aligned}$$

where $C$ is the constant of integration. $_\square$

Evaluate $\displaystyle\int \arcsin 3x \,dx.$

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Let $u=\arcsin 3x$ and $dv=dx,$ then $du = \frac{3\, dx}{\sqrt{1-(3x)^{2}}}$ and $v=x,$ which implies that the given expression is

$$\displaystyle\int \arcsin 3x \,dx = \displaystyle x \arcsin 3x - \int \dfrac{3x}{\sqrt{1-9x^{2}}} \, dx. \qquad (1)$$

Let $a= 1-9x^{2}$ and $\frac{da}{-18x} = dx ,$ then $(1)$ is equivalent to

$$\begin{aligned} \displaystyle x \arcsin 3x + \dfrac{1}{6} \int \dfrac{da}{\sqrt{a}} &= \displaystyle x \arcsin 3x + \dfrac{1}{3} \sqrt{a} + C \\ & = \displaystyle x \arcsin 3x + \dfrac{1}{3} \sqrt{1-9x^{2}} + C, \end{aligned}$$

where $C$ is the constant of integration. $_\square$

Evaluate $\displaystyle \int e^{2x}\sin x \,dx.$

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Let $f(x)=\sin x$ and $g''(x)=e^{2x},$ so that $f'(x)=\cos x,~f''(x)=-\sin x,~g'(x)=\dfrac{1}{2}e^{2x},$ and $g(x)=\dfrac{1}{4}e^{2x}.$ Then

$$\begin{aligned} \int e^{2x}\sin x \,dx &= \int f(x)g''(x) \,dx \\ &= f(x)g'(x) - \int f'(x)g'(x) \,dx \\ &= f(x)g'(x) - \left( f'(x)g(x) - \int f''(x)g(x) \,dx\right) \\ &= \dfrac{1}{2}e^{2x}\sin x - \dfrac{1}{4}e^{2x}\cos x + \int -\dfrac{1}{4}e^{2x}\sin x \,dx + c. \end{aligned}$$

Let $\displaystyle \text{Fi}(x)=\int e^{2x}\sin x \,dx.$ Then we see that

$$\begin{aligned} \text{Fi}(x) &= \dfrac{1}{2}e^{2x}\sin x - \dfrac{1}{4}e^{2x}\cos x - \dfrac{1}{4}\text{Fi}(x) + c \\ \dfrac{5}{4}\text{Fi}(x) &= \dfrac{1}{4}e^{2x}(2\sin x-\cos x) + c \\ \text{Fi}(x) &=\dfrac{e^{2x}}{5}(2\sin x-\cos x) + C, \end{aligned}$$

where $C$ is the constant of integration. $_\square$

• $u$-Substitution
• Integration Techniques