Integration by Parts Formula

\[ \boxed{\bbox[#FAF8ED,5pt]{ \int u(x) v'(x)\,dx = u(x)v(x) - \int u'(x)v(x)\,dx}}\tag{1} \]

The Integration by Parts formula is often written using differentials: \[ \int u\,dv = uv - \int v\,du \]

where \(dv = v'(x)\,dx\) and \(du = u'(x)\,dx\).

EXAMPLE 1

Evaluate \(\displaystyle\int x~\cos x\,dx\)

Solution The integrand is a product, so we try writing \(x~\cos x = uv'\) with \[ u(x) = x,\qquad v'(x) = \cos x \]

In this case, \(u'(x) = 1\) and \(v(x) = \sin x\). By the Integration by Parts formula, \[ \int \underbrace{x~\cos x}_{uv'}\,dx = \underbrace{x~\sin x}_{uv} - \int \underbrace{\sin x}_{u'v}\,dx = x~\sin x + \cos x + C \]

Let's check the answer by taking the derivative: \[ {\frac{d}{dx}(x~\sin x + \cos x + C) = x~\cos x + \sin x - \sin x = x~\cos x} \]

In applying Eq.(1), any antiderivative \(v(x)\) of \(v'(x)\) may be used.

EXAMPLE 2 Good Versus Bad Choices of \(u\) and \(v'\)

Evaluate \(\displaystyle\int x e^x \,dx\)

Solution Based on our guidelines, it makes sense to write \(x e^x = uv'\) with

• \(u=x\) (since \(u'=1\) is simpler)
• \(v'=e^x\) (since we can evaluate \({v=\int e^x\,dx = e^x + C)}\)

Integration by Parts gives us \[ \int xe^x \,dx = u(x)v(x) - \int u'(x) v(x)\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C \]

Let's see what happens if we write \(xe^x =uv'\) with \(u = e^x\), \(v'= x\). Then \[ \begin{eqnarray*} &\displaystyle u'(x) = e^x,\qquad v(x) = \int x\,dx = \frac12x^2 + C\\ &\displaystyle \int \underbrace{x e^x}_{uv'} \,dx = \underbrace{\dfrac12x^2e^x}_{uv}\,\, -\,\, \int\underbrace{ \dfrac12x^2e^x}_{u'v}dx \end{eqnarray*} \]

This is a poor choice of \(u\) and \(v'\) because the integral on the right is more complicated than our original integral.

In Example 3, it makes sense to take \(u=x^2\) because Integration by Parts reduces the integration of \(x^2\cos x\) to the integration of \(2x\sin x\), which is easier.

EXAMPLE 3 Integrating by Parts More Than Once

Evaluate \(\int x^2~\cos x \,dx\).

Solution Apply Integration by Parts a first time with \(u = x^2\) and \(v' = \cos x\): \[ {\int \underbrace{x^2~\cos x}_{uv'}\, dx = \underbrace{x^2~\sin x}_{uv} - \int \underbrace{ 2x\,\sin x }_{u'v}\,dx = x^2~\sin x - 2\int x~\sin x\,dx}\tag{2} \]

Now apply it again to the integral on the right, this time with \(u=x\) and \(v'=\sin x\): \[ \int \underbrace{x~\sin x}_{uv'}\, dx = \underbrace{-x~\cos x}_{uv} - \int \underbrace{ (-\cos x)}_{u'v}\,dx = -x~\cos x+\sin x + C \]

Using this result in Eq.(2), we obtain \[ \begin{align*} \int x^2~\cos x\, dx & = x^2~\sin x - 2\int x~\sin x\,dx = x^2~\sin x - 2(-x~\cos x+\sin x)+C\\ &= x^2~\sin x +2x~\cos x-2\sin x+C \end{align*} \]

Surprisingly, the choice \(v'=1\) is effective in some cases. Using it as in Example 4, we find that \[ \int \ln x\,dx = x~\ln x - x + C \] This choice also works for the inverse trigonometric functions (see Exercise 6).

EXAMPLE 4 Taking \(v'=1\)

Evaluate \(\int_1^3 \ln x \,dx\)

Solution The integrand is not a product, so at first glance, this integral does not look like a candidate for Integration by Parts. However, we are free to add a factor of \(1\) and write \(\ln x = (\ln x)\cdot 1 = uv'\). Then \[ \begin{alignat*}{2} u &= \ln x,\qquad &v' &= 1\\ 'u &= x^{-1},\qquad &v &= x \end{alignat*} \] \[ \int_1^3 \underbrace{ \ln x }_{uv'} \,dx = \underbrace{ x\ln x}_{uv}\bigg|_1^3 - \int_1^3 \underbrace{ 1}_{u'v}\, dx = (3\ln 3-0) - 2 = 3\ln 3 -2 \]

In Example 5, the choice \(u = e^x\), \(v'= \cos x\) works equally well.

EXAMPLE 5 Going in a Circle?

Evaluate \(\int e^x~\cos x \,dx\).

Solution There are two reasonable ways of writing \( e^x~\cos x\) as \(uv'\). Let's try \(u=\cos x\) and \(v'= e^x\). Then \[ \int \underbrace{e^x~\cos x }_{uv'} \,dx = \underbrace{e^x~\cos x}_{uv} + \int \underbrace{ e^x~\sin x}_{-u'v} \, dx\tag{3}\]

Now use Integration by Parts to the integral on the right with \(u= \sin x\) and \(v'=e^x\): \[\int e^x~\sin x \,dx = e^x~\sin x - \int e^x~\cos x \, dx\tag{4}\]

Eq.(4) brings us back to our original integral of \(e^x~\cos x\), so it looks as if we're going in a circle. But we can substitute Eq.(4) in Eq.(3) and solve for the integral of \(e^x~\cos x\): \[ \begin{align*} \int e^x~\cos x \,dx &= e^x~\cos x + \int e^x~\sin x \, dx = e^x~\cos x + e^x~\sin x - \int e^x~\cos x \, dx\\ 2\int e^x~\cos x\,dx &= e^x~\cos x + e^x~\sin x + C\\ \int e^x~\cos x \,dx &= \frac{1}{2}e^x(\cos x + \sin x) + C \end{align*} \]

A reduction formula (also called a recursive formula) expresses the integral for a given value of \(n\) in terms of a similar integral for a smaller value of \(n\). The desired integral is evaluated by applying the reduction formula repeatedly.

In general, \(\int x^ne^x\,dx=P_n(x)e^x + C\), where \(P_n(x)\) is a polynomial of degree \(n\) (see Exercise 78).

EXAMPLE 6 A Reduction Formula

Derive the reduction formula \[ \boxed{\bbox[#FAF8ED,5pt]{ \int x^n e^x \,dx = x^n e^x - n \int x^{n-1} e^x\, dx}}\tag{5} \]

Then evaluate \(\displaystyle\int x^3 e^x \,dx\).

Solution We apply Integration by Parts with \(u=x^n\) and \(v'=e^x\): \[ \int x^n e^x \,dx = uv-\int u'v\,dx = x^n e^x - n\int x^{n-1} e^x\, dx \]

To evaluate \(\displaystyle\int x^3 e^x\, dx\), we'll need to use the reduction formula for \(n = 3, 2, 1\):

\[ \begin{align*} \int x^3 e^x\, dx &= x^3 e^x - 3\int x^2 e^x\, dx\\ & = x^3 e^x - 3\left(x^2 e^x - 2\int x e^x\, dx\right)\\ & = x^3 e^x - 3 x^2 e^x +6\int x e^x\, dx\\ & = x^3 e^x - 3 x^2 e^x +6\left(x e^x - \int e^x\, dx\right)\\ & = x^3 e^x - 3 x^2 e^x +6 x e^x - 6 e^x+C\\ &=(x^3-3x^2+6x-6)e^x +C \end{align*} \]