Integration by parts

Integration by parts (IBP) is a method of integration with the formula: Template:Align

Or more compactly, Template:Align

where $u$ and $v$ are functions of a variable, for instance, $x$, giving $u(x)$ and $v(x)$.

ILATE Rule

A rule of thumb has been proposed, consisting of choosing $u$ as the function that comes first in the following list:

I – inverse trigonometric functions: $\arctan (x),\mathrm{arcsec}(x),$ etc.

L – logarithmic functions: $\ln (x),{\log }_b(x),$ etc.

A – polynomials: $x^2,3x^{50},$ etc.

T – trigonometric functions: $\sin (x),\tan (x),$ etc.

E – exponential functions: $e^x,19^x,$ etc.

Derivation

The theorem can be derived as follows. For two continuously differentiable functions $u(x)$ and $v(x)$, the product rule states:

$(u(x)v(x){)}^{\prime }=v(x)u^{\prime }(x)+u(x)v^{\prime }(x).$

Integrating both sides with respect to $x$,

$\int (u(x)v(x){)}^{\prime }dx=\int u^{\prime }(x)v(x)dx+\int u(x)v^{\prime }(x)dx,$

and noting that an indefinite integral is an antiderivative gives

$u(x)v(x)=\int u^{\prime }(x)v(x)dx+\int u(x)v^{\prime }(x)dx,$

where we neglect writing the constant of integration. This yields the formula for integration by parts:

$\int u(x)v^{\prime }(x)dx=u(x)v(x)-\int u^{\prime }(x)v(x)dx,$

or in terms of the differentials $du=u^{\prime }(x)dx,dv=v^{\prime }(x)dx,$

$\int u(x)dv=u(x)v(x)-\int v(x)du.$

This is to be understood as an equality of functions with an unspecified constant added to each side. Taking the difference of each side between two values x = a and x = b and applying the fundamental theorem of calculus gives the definite integral version:

$\underset{a}{\overset{b}{\int }}u(x)v^{\prime }(x)dx=u(b)v(b)-u(a)v(a)-\underset{a}{\overset{b}{\int }}{u}^{\prime }(x)v(x)dx.$

Examples

Functions multiplied by one and itself

Given $I=\int \ln (x)\cdot 1dx$

The first example is ∫ ln(x) dx. We write this as:

$I=\int \ln (x)\cdot 1dx.$

Let:

$u=\ln (x)\Rightarrow du=\frac{dx}{x}$

$dv=dx\Rightarrow v=x$

then:

$\begin{array}{cc}\int \ln (x)dx& =x\ln (x)-\int \frac{x}{x}dx\\ & =x\ln (x)-\int 1dx\\ & =x\ln (x)-x+C\end{array}$

where C is the constant of integration.

Given $I=\int \arctan (x)dx.$

The second example is the inverse tangent function arctan(x):

$I=\int \arctan (x)dx.$

Rewrite this as

$\int \arctan (x)\cdot 1dx.$

Now let:

$u=\arctan (x)\Rightarrow du=\frac{dx}{1+x^2}$

$dv=dx\Rightarrow v=x$

then

$\begin{array}{cc}\int \arctan (x)dx& =x\arctan (x)-\int \frac{x}{1+x^2}dx\\ & =x\arctan (x)-\frac{\ln (1+x^2)}{2}+C\end{array}$

using a combination of the inverse chain rule method and the natural logarithm integral condition.

Polynomials and trigonometric functions

In order to calculate

$I=\int x\cos (x)dx,$

let:

$u=x\Rightarrow du=dx$

$dv=\cos (x)dx\Rightarrow v=\int \cos (x)dx=\sin (x)$

then:

$\begin{array}{cc}\int x\cos (x)dx& =\int udv\\ & =u\cdot v-\int vdu\\ & =x\sin (x)-\int \sin (x)dx\\ & =x\sin (x)+\cos (x)+C,\end{array}$

where C is a constant of integration.

For higher powers of x in the form

$\int x^n{e}^{x}dx,\int x^n\sin (x)dx,\int x^n\cos (x)dx,$

repeatedly using integration by parts can evaluate integrals such as these; each application of the theorem lowers the power of x by one.

Exception to LIATE

$\int x^3{e}^{x^2}dx,$

one would set

$u=x^2,dv=x\cdot e^{x^2}dx,$

so that

$du=2xdx,v=\frac{e^{x^2}}{2}.$

Then

$\int x^3{e}^{x^2}dx=\int (x^2)\left(x{e}^{x^2}\right)dx=\int udv=uv-\int vdu=\frac{x^2{e}^{x^2}}{2}-\int x{e}^{x^2}dx.$

Finally, this results in

$\int x^3{e}^{x^2}dx=\frac{e^{x^2}(x^2-1)}{2}+C.$

Performing IBP twice

$I=\int e^x\cos (x)dx.$

Here, integration by parts is performed twice. First let

$u=\cos (x)\Rightarrow du=-\sin (x)dx$

$dv=e^{x}dx\Rightarrow v=\int e^{x}dx=e^x$

then:

$\int e^x\cos (x)dx=e^x\cos (x)+\int e^x\sin (x)dx.$

Now, to evaluate the remaining integral, we use integration by parts again, with:

$u=\sin (x)\Rightarrow du=\cos (x)dx$

$dv=e^{x}dx\Rightarrow v=\int e^{x}dx=e^x.$

Then:

$\int e^x\sin (x)dx=e^x\sin (x)-\int e^x\cos (x)dx.$

Putting these together,

$\int e^x\cos (x)dx=e^x\cos (x)+e^x\sin (x)-\int e^x\cos (x)dx.$

The same integral shows up on both sides of this equation. The integral can simply be added to both sides to get

$2\int e^x\cos (x)dx=e^x[\sin (x)+\cos (x)]+C,$

which rearranges to

$\int e^x\cos (x)dx=\frac{1}{2}{e}^x[\sin (x)+\cos (x)]+C^{\prime }$

Problem Set

1) $\int 2x\cos (5-13x)dx$

2) $\int x\cos (x)dx$

3) $\int x^3\sin (3x)dx$

4) $\int e^{2x}\cos (3x)dx$

5) $\int e^{-x}(2x^2-3x+5)dx$

6) $\int 4{\tan }^{-1}(\frac{3}{x})dx$

7) $\int x\tan (x-4x^2)dx$

8) $\int x^5{\cos }^{-1}(3x)dx$

9) $\int e^{2x}3x^{-2}dx$

10) $\int x\ln (3x)dx$