1.8 Integration by parts
Integrals like
∫ x e − 6 x d x \int x e^{-6 x} \d{x} ∫ x e − 6 x d x are not solvable by substitution , but they look
like they are related closely enough that you may be able to! What to do here?
Recall the product rule for two functions u and v ,
then integrate:
( u v ) ′ = u v ′ + v u ′ u v = ∫ u d v + ∫ v d u \begin{align}
(uv)' &= uv' + v u' \\
uv &= \int u \d{v} + \int v \d{u}
\end{align} ( uv ) ′ uv = u v ′ + v u ′ = ∫ u d v + ∫ v d u Rearranging gives the familiar form:
Let’s try this for the integral above with
u = x d v = e − 6 x d x d u = d x v = − e − 6 x 6 \begin{align}
u &= x & \d{v} &= e^{-6x} \d{x} \\
\d{u} &= \d{x} & v &= -\frac{e^{-6x}}{6}
\end{align} u d u = x = d x d v v = e − 6 x d x = − 6 e − 6 x so
∫ x e − 6 x d x = x − e − 6 x 6 − ∫ − 1 6 e − 6 x d x = − x e − 6 x 6 + 1 6 ∫ e − 6 x d x = − x e − 6 x 6 − e − 6 x 36 + c \begin{align}
\int x e^{-6x} \d{x}
&= x\frac{-e^{-6x}}{6} - \int\frac{-1}{6} e^{-6x} \d{x} \\
&= -\frac{xe^{-6x}}{6} + \frac{1}{6} \int e^{-6x} \d{x} \\
&= -\frac{xe^{-6x}}{6} - \frac{e^{-6x}}{36} + c
\end{align} ∫ x e − 6 x d x = x 6 − e − 6 x − ∫ 6 − 1 e − 6 x d x = − 6 x e − 6 x + 6 1 ∫ e − 6 x d x = − 6 x e − 6 x − 36 e − 6 x + c Choosing parts ¶ The heart of the technique is making a suitable choice for u and d v \d{v} d v
Example 1 ¶ Evaluate
∫ ( 3 x + 5 ) cos ( x 4 ) d x \int(3x+5)\cos\left(\frac{x}{4}\right) \d{x} ∫ ( 3 x + 5 ) cos ( 4 x ) d x Let
u = 3 x + 5 d v = cos ( x 4 ) d x d u = 3 d x v = 4 sin ( x 4 ) \begin{align}
u &= 3x+5 & \d{v} &= \cos\left(\frac{x}{4}\right) \d{x} \\
\d{u} &= 3 \d{x} & v &= 4\sin\left(\frac{x}{4}\right)
\end{align} u d u = 3 x + 5 = 3 d x d v v = cos ( 4 x ) d x = 4 sin ( 4 x ) so:
∫ ( 3 x + 5 ) cos ( x 4 ) d x = ( 3 x + 5 ) [ 4 sin ( x 4 ) ] − ∫ 4 sin ( x 4 ) 3 d x = ( 12 x + 20 ) sin ( x 4 ) − 12 ∫ sin ( x 4 ) d x = ( 12 x + 20 ) sin ( x 4 ) + 48 cos ( x 4 ) + c \begin{align}
\int(3x+5)\cos\left(\frac{x}{4}\right) \d{x}
&= (3x+5)\left[4\sin\left(\frac{x}{4}\right)\right] -
\int 4 \sin \left(\frac{x}{4}\right) 3 \d{x} \\
&= (12x+20) \sin\left(\frac{x}{4}\right) -
12\int\sin\left(\frac{x}{4}\right) \d{x} \\
&= (12x+20) \sin\left(\frac{x}{4}\right) + 48\cos\left(\frac{x}{4}\right) + c
\end{align} ∫ ( 3 x + 5 ) cos ( 4 x ) d x = ( 3 x + 5 ) [ 4 sin ( 4 x ) ] − ∫ 4 sin ( 4 x ) 3 d x = ( 12 x + 20 ) sin ( 4 x ) − 12 ∫ sin ( 4 x ) d x = ( 12 x + 20 ) sin ( 4 x ) + 48 cos ( 4 x ) + c Example 2 ¶ Evaluate
∫ ln x d x \int \ln x \d{x} ∫ ln x d x Let
u = ln x d v = d x d u = 1 x d x v = x \begin{align}
u &= \ln x & \d{v} &= \d{x} \\
\d{u} &= \frac{1}{x} \d{x} & v &= x
\end{align} u d u = ln x = x 1 d x d v v = d x = x so:
∫ ln x d x = x ln x − ∫ 1 x x d x = x ln x − ∫ d x = x ln x − x + c \begin{align}
\int \ln x \d{x} &= x \ln x - \int\frac{1}{x} x \d{x} \\
&= x \ln x - \int \d{x} \\
&= x \ln x - x + c \\
\end{align} ∫ ln x d x = x ln x − ∫ x 1 x d x = x ln x − ∫ d x = x ln x − x + c Example 3 ¶ Evaluate
∫ x 5 x 3 + 1 d x \int x^5 \sqrt{x^3+1} \d{x} ∫ x 5 x 3 + 1 d x Let
u = x 3 d v = x 2 x 3 + 1 d x d u = 3 x 2 d x v = 2 9 ( x 3 + 1 ) 3 / 2 \begin{align}
u &= x^3 & \d{v} &= x^2\sqrt{x^3+1} \d{x} \\
\d{u} &= 3x^2 \d{x} & v &= \frac{2}{9}(x^3+1)^{3/2}
\end{align} u d u = x 3 = 3 x 2 d x d v v = x 2 x 3 + 1 d x = 9 2 ( x 3 + 1 ) 3/2 so:
∫ ( x 5 ) ( ( x 3 + 1 ) ) d x = x 3 2 9 ( x 3 + 1 ) 3 / 2 − ∫ [ 2 9 ( x 3 + 1 ) 3 / 2 ] 3 x 2 d x = 2 9 x 3 ( x 3 + 1 ) 3 / 2 − 2 9 ⋅ 2 5 ( x 3 + 1 ) 5 / 2 + c = 2 9 x 3 ( x 3 + 1 ) 3 / 2 − 4 45 ( x 3 + 1 ) 5 / 2 + c \begin{align}
\int(x^5)(\sqrt(x^3+1))dx
&= x^3\frac{2}{9}(x^3+1)^{3/2} -
\int\left[\frac{2}{9}(x^3+1)^{3/2}\right]3x^2dx \\
&= \frac{2}{9}x^3(x^3+1)^{3/2} -
\frac{2}{9}\cdot\frac{2}{5}(x^3+1)^{5/2} + c \\
&= \frac{2}{9}x^3(x^3+1)^{3/2} - \frac{4}{45}(x^3+1)^{5/2} + c
\end{align} ∫ ( x 5 ) ( ( x 3 + 1 )) d x = x 3 9 2 ( x 3 + 1 ) 3/2 − ∫ [ 9 2 ( x 3 + 1 ) 3/2 ] 3 x 2 d x = 9 2 x 3 ( x 3 + 1 ) 3/2 − 9 2 ⋅ 5 2 ( x 3 + 1 ) 5/2 + c = 9 2 x 3 ( x 3 + 1 ) 3/2 − 45 4 ( x 3 + 1 ) 5/2 + c Tabular method ¶ To integrate
∫ x 4 e x / 2 d x \int x^4 e^{x/2} \d{x} ∫ x 4 e x /2 d x by parts, let
u = x 4 d v = e x / 2 d u = 4 x 3 d x v = 2 e x / 2 \begin{align}
u &= x^4 & \d{v} &= e^{x/2} \\
\d{u} &= 4x^3 \d{x} & v &= 2e^{x/2}
\end{align} u d u = x 4 = 4 x 3 d x d v v = e x /2 = 2 e x /2 which allows:
∫ x 4 e x / 2 d x = 2 x 4 e x / 2 − ∫ 8 x 3 e x / 2 d x \int x^4e^{x/2} \d{x} = 2 x^4 e^{x/2} - \int 8 x^3 e^{x/2} \d{x} ∫ x 4 e x /2 d x = 2 x 4 e x /2 − ∫ 8 x 3 e x /2 d x We will need to integrate by parts again, and again, and again... but there is
a shortcut! Make a staggered table. Differentiate u repeatedly until you get
a 0 in the last row. Then, integrate d v \d{v} d v u . Multiply across rows using opposite signs.
sign u u u d v \d{v} d v e x / 2 e^{x/2} e x /2 + + + x 4 x^4 x 4 2 e x / 2 2e^{x/2} 2 e x /2 − - − 4 x 3 4x^3 4 x 3 4 e x / 2 4e^{x/2} 4 e x /2 + + + 12 x 2 12x^2 12 x 2 8 e x / 2 8e^{x/2} 8 e x /2 − - − 24 x 24x 24 x 16 e x / 2 16e^{x/2} 16 e x /2 + + + 24 32 e x / 2 32e^{x/2} 32 e x /2 0
The result is:
∫ x 4 e x / 2 d x = x 4 ( 2 e x / 2 ) − 4 x 3 ( 4 e x / 2 ) + 12 x 2 ( 8 e x / 2 ) − 24 x ( 16 e x / 2 ) + 24 ( 32 e x / 2 ) = ( 2 x 4 − 16 x 3 + 96 x 2 − 384 x + 768 ) e x / 2 \begin{align}
\int x^4 e^{x/2} \d{x} &= x^4 (2e^{x/2}) - 4x^3 (4e^{x/2}) \\
&+ 12x^2 (8e^{x/2}) - 24x (16e^{x/2}) + 24 (32e^{x/2}) \\
&= (2x^4 - 16x^3 + 96x^2 - 384x + 768) e^{x/2}
\end{align} ∫ x 4 e x /2 d x = x 4 ( 2 e x /2 ) − 4 x 3 ( 4 e x /2 ) + 12 x 2 ( 8 e x /2 ) − 24 x ( 16 e x /2 ) + 24 ( 32 e x /2 ) = ( 2 x 4 − 16 x 3 + 96 x 2 − 384 x + 768 ) e x /2 Skill builder problems ¶ Evaluate the following
Select the following parts:
u = ln x d v = ( x + 2 ) d x d u = 1 x d x v = x 2 2 + 2 x \begin{align}
u &= \ln x & \d{v} &= (x+2)\d{x} \\
\d{u} &= \frac{1}{x}\d{x} & v &= \frac{x^2}{2} + 2x
\end{align} u d u = ln x = x 1 d x d v v = ( x + 2 ) d x = 2 x 2 + 2 x so
∫ ( x + 2 ) ln x d x = ( x 2 2 + 2 x ) ln x − ∫ ( x 2 2 + 2 x ) 1 x d x = ( x 2 2 + 2 x ) ln x − ∫ ( x 2 + 2 ) d x = ( x 2 2 + 2 x ) ln x − x 2 4 − 2 x + c \begin{align}
\int (x+2)\ln x \d{x} &= \left( \frac{x^2}{2} + 2x \right)\ln x -
\int \left( \frac{x^2}{2} + 2x \right) \frac{1}{x} \d{x} \\
&= \left( \frac{x^2}{2} + 2x \right)\ln x -
\int \left( \frac{x}{2} + 2 \right) \d{x} \\
&= \left( \frac{x^2}{2} + 2x \right)\ln x - \frac{x^2}{4} - 2x + c
\end{align} ∫ ( x + 2 ) ln x d x = ( 2 x 2 + 2 x ) ln x − ∫ ( 2 x 2 + 2 x ) x 1 d x = ( 2 x 2 + 2 x ) ln x − ∫ ( 2 x + 2 ) d x = ( 2 x 2 + 2 x ) ln x − 4 x 2 − 2 x + c Since x 5 x^5 x 5
sign u u u d v \d{v} d v sin x \sin x sin x + + + x 5 x^5 x 5 − cos x -\cos x − cos x − - − 5 x 4 5x^4 5 x 4 − sin x -\sin x − sin x + + + 20 x 3 20x^3 20 x 3 cos x \cos x cos x − - − 60 x 2 60x^2 60 x 2 sin x \sin x sin x + + + 120 x 120x 120 x − cos x -\cos x − cos x − - − 120 − sin x -\sin x − sin x 0
so
∫ x 5 sin x d x = − x 5 cos x + 5 x 4 sin x + 20 x 3 cos x − 60 x 2 sin x − 120 x cos x + 120 sin x + c \int x^5\sin x \d{x} = -x^5 \cos x + 5x^4 \sin x + 20x^3 \cos x -
60x^2 \sin x - 120x \cos x + 120 \sin x + c ∫ x 5 sin x d x = − x 5 cos x + 5 x 4 sin x + 20 x 3 cos x − 60 x 2 sin x − 120 x cos x + 120 sin x + c To make the solution look cleaner, you can factor out cos x \cos x cos x s i n x \ sin x s in x
∫ x 5 sin x d x = ( − x 5 + 20 x 3 − 120 x ) cos x + ( 5 x 4 − 60 x 2 + 120 ) sin x + c \int x^5\sin x \d{x} = (-x^5 + 20x^3 - 120x) \cos x +
(5x^4 - 60x^2 + 120) \sin x + c ∫ x 5 sin x d x = ( − x 5 + 20 x 3 − 120 x ) cos x + ( 5 x 4 − 60 x 2 + 120 ) sin x + c Select the following parts:
u = cos x d v = e x d x d u = − sin x d x v = e x \begin{align}
u &= \cos x & \d{v} &= e^x\d{x} \\
\d{u} &= -\sin x\d{x} & v &= e^x
\end{align} u d u = cos x = − sin x d x d v v = e x d x = e x so
∫ e x cos x d x = e x cos x + ∫ e x sin x d x \int e^x \cos x \d{x} = e^x \cos x + \int e^x \sin x \d{x} ∫ e x cos x d x = e x cos x + ∫ e x sin x d x To evaluate this integral, select new parts:
u = sin x , d v = e x d x d u = cos x d x v = e x \begin{align}
u &= \sin x, & \d{v} &= e^x\d{x} \\
\d{u} &= \cos x\d{x} & v &= e^x
\end{align} u d u = sin x , = cos x d x d v v = e x d x = e x so
∫ e x cos x d x = e x cos x + [ e x sin x − ∫ e x cos x d x ] \int e^x \cos x \d{x} = e^x \cos x+ \left[e^x \sin x - \int e^x \cos x \d{x}\right] ∫ e x cos x d x = e x cos x + [ e x sin x − ∫ e x cos x d x ] The same integral appears on both sides! Add them together and solve:
2 ∫ e x cos x d x = e x cos x + e x sin x ∫ e x cos x d x = e x 2 ( cos x + sin x ) + c \begin{align}
2 \int e^x \cos x \d{x} &= e^x \cos x + e^x \sin x \\
\int e^x \cos x \d{x} &= \frac{e^x}{2}(\cos x + \sin x) + c
\end{align} 2 ∫ e x cos x d x ∫ e x cos x d x = e x cos x + e x sin x = 2 e x ( cos x + sin x ) + c