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

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

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

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

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

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

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:

\[\begin{align} &&\text{Logs} &&\text{Polynomials} &&\text{Trigs} &&\text{Exponentials}\\ &&u\ \ \ \ &&\cdots\ \ \ \ \ \ &&\cdots\ \ &&v'\ \ \ \ \ \ \end{align}\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

where \(C\) is the constant of integration. \(_\square\)

• \(u\)-Substitution
• Integration Techniques