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Integration by Parts

April 5th, 2009 |

Author: Merrin

Integration by parts is analogous to the product rule for differentiation. If the integrand consists of the product of two functions then integration by parts may help. Sometimes one of these functions is just the plain function f(x)=1 so don’t ignore the trivial case.

Theorem 1 Integration by parts.

First form

$\cfrac{d}{dx} \left[f(x)g(x)\right] = f'(x)g(x) + f(x)g'(x)$

$f(x)g(x) =\displaystyle \int f'(x)g(x)gdx+\displaystyle \int f(x)g'(x) dx$

$\displaystyle \int f(x)g'(x)dx=f(x)g(x)-\displaystyle \int f'(x)g(x) dx$

Second form:

$(uv)'=u'v+uv'$

$d(uv)=vdu+udv$

$\displaystyle \int udv =\displaystyle \int d(uv) -\displaystyle \int vdu$

$\displaystyle \int udv=uv-\displaystyle \int vdu$

Amongst the two integrals there is a trading of one derivative and one antiderivative. Doing integration by parts is similar for definite integrals also.

Theorem 2. Integration by parts for definite integrals

$\displaystyle \int _{a}^{b}f(x)\cfrac{dg(x)}{dx} dx=\left. f(x)g(x)\right|_{x=a}^{x=b} -\displaystyle \int _{a}^{b}\cfrac{df(x)}{dx} g(x)dx$

Example 1. Evaluate the integral by integration by parts.

$\displaystyle \int x\text{cos}\,\, xdx$

Solution 1. When integrating by parts it is often helpful to make a small table of all the pieces in the equation.

$\displaystyle \int x\text{cos}\,\, xdx$
$u=x\quad dv=\text{cos}\,\, xdx\quad v=\text{sin}\,\, x\quad du=dx$
$\displaystyle \int x\text{cos}\,\, xdx =x\text{sin}\,\, x-\displaystyle \int \text{sin}\,\, xdx =x\text{sin}\,\, x+\text{cos}\,\, x+C$

Example 2. Evaluate the following integral by integration by parts.

$\displaystyle \int xe^{x} dx$

Solution 2. Again we make a table. Choose u and v wisely otherwise the wrong substitution makes things worse.

$\displaystyle \int xe^{x} dx$
$u=e^{x}\quad dv=xdx\quad v=x^{2} /2\quad du=e^{x} dx$

This is the wrong substitution, v is growing in complexity.

$\int udv= \, uv-\displaystyle \int vdu$
$\int xe^{x} dx =\cfrac{x^{2} e^{x} }{2} -\cfrac{1}{2} \displaystyle \int x^{2} e^{x} dx =?$

Clearly, we have made the wrong substitution and the integrand we are left with has grown more monstrous. Let us make the right choices and solve the integral.

$\int xe^{x} dx$
${u=x,\quad \, dv=e^{x} dx,\quad \, v=e^{x} ,\quad \, du=dx, }$
$\int udv= \, uv-\displaystyle \int vdu$
$\int xe^{x} dx =xe^{x} -\displaystyle \int e^{x} dx =xe^{x} -e^{x} +C$

Example 3. Another integral like this one that can be solved by integration by parts is

$\displaystyle \int \ln xdx$

But where is the second function in the integrand?

Solution 3. The second function is just one. We integrate by parts once to get

$\int \ln xdx$
$u=\ln x \quad dv=1dx\quad v=x\quad du=\cfrac{1}{x} dx$
$\displaystyle \int udv= uv-\displaystyle \int vdu$
$\int \ln xdx =x\ln x-\displaystyle \int x\cfrac{1}{x} dx =x\ln x-x+C$

Example 4. Calculate the following integral by integrating by parts twice and solving algebraically for the integral.

$I=\displaystyle \int e^{x} \text{cos}\,\, xdx$

Solution 4. By now we know how to integrate by parts, so I apply the procedure twice suppressing the details. The interesting part of this integral is that by integrating by parts twice a piece of the new equation is what

we started with. One way to represent the integral is with capital I. To finish, solve for I algebraically.

$I = {\displaystyle \int e^{x} \text{cos}\,\, xdx=e^{x} \text{sin}\,\, x-\displaystyle \int e^{x} \text{sin}\,\, xdx}$