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Integration by Substitution

April 5th, 2009 |

Author: Merrin

Integration by substitution is a change of variable procedure that transforms integrals. In terms of new variables, the integral is simplified or has a known solution. Knowing what to substitute is a bit of an art. One thing to look for is the derivative of the substitution alone in the integrand.

Theorem 1. Integration by substitution, indefinite integrals.

$F(x)=\displaystyle \int f(u)\cfrac{du}{dx} dx=\displaystyle \int f(u)du$

You may be wondering how the last integral can be a function of x when only u appears. The symbol u is just a dummy variable. Once the integral is transformed it is possible to just use x again but it is not very clear so u is used. In the above form, we see a function of u and the derivative of u. The justification for the transformation is the fundamental theorem of calculus using as an integrand,

$[F(u(x))]'=f(u(x))u'(x)$

This derivative is calculated using the chain rule. The symbol u is not required but is commonly used.

Theorem 2. Integration by substitution, definite integrals.

$\displaystyle \int _{a}^{b}f(u(x))\cfrac{du}{dx} dx=\displaystyle \int _{u(a)}^{u(b)}f(u)du =F(u(b))-F(u(a))$

The formula for definite integrals is the same, but the limits of integration generally change. Correct the limits of integration by evaluating u at the old limits.

Example 1. Use u substitution to calculate the following indefinite integral

$\displaystyle \int \text{sin}\,\, (ax)dx$

Solution 1. For functions of the type f(ax) try a substitution of u=ax which is a scale transformation. The function we are transforming is ax and its derivative is proportional to one which is also present in the integrand.

$u=ax \quad \cfrac{du}{dx} =a\to du=adx$

$\int \text{sin}\,\, (ax)dx=\cfrac{1}{a} \displaystyle \int \text{sin}\,\, udu =-\cfrac{1}{a} \text{cos}\,\, u+C=-\cfrac{1}{a} \text{cos}\,\, (ax)+C$

Alternate Solution: I like to set up the expression so I can do some of the work in my head and not have to write u every time and then write its derivative. Take any object inside parenthesis to be a blob or its own variable.

$d(ax)=xd(a)+ad(x)=adx$

$\displaystyle \int \text{sin}\,\, (ax)dx =\displaystyle \int \text{sin}\,\, (ax)\cfrac{d(ax)}{a} =\cfrac{1}{a} \displaystyle \int \text{sin}\,\, (ax)d(ax)$

$I =-\cfrac{1}{a} \text{cos}\,\, (ax)+C$

Doing it this way, we don’t have to write down u equals such and such then find du in terms of dx. It is however a good idea to write everything out when first learning so as to check that all the steps are correct.

Example 2. Here is an integral that resists integration.

$\displaystyle \int e^{e^{x} } dx=?$

The solution to this integral in terms of elementary functions is not known. Let us look what happens when we add more stuff to the integrand.

$\displaystyle \int e^{e^{x} } e^{x} dx =?$

Solution 2. Since we are familiar with the chain rule we can recognize the integrand as the derivative of an iterated exponential. However, we will find the antiderivative by the method of substitution.

${u=e^{x} \quad \cfrac{du}{dx} =e^{x} \to du=e^{x} dx} \\ {\displaystyle \int e^{e^{x} } e^{x} dx\to \displaystyle \int e^{u} du=e^{u} +C=e^{e^{x} } +C }$

See, that integral wasn’t really that tough. Solve for the antiderivative in terms of u but put back the expression for u(x) in terms of x.

Example 3. Let us continue the game of spotting a function and its derivative.

Calculate the indefinite integral of

$\displaystyle \int \cfrac{dx}{(1+x^{2} )\text{arctan}\,\, x}$

Solution 3. Wise substitution choices simplify integrands while others complicate matters. We see that

$(1+x^{2} )^{-1}$

is the derivative of arctanx so this is a good choice for substitution.

${u=\text{arctan}\,\, x\qquad \cfrac{du}{dx} =\cfrac{1}{1+x^{2} } \to du=\cfrac{dx}{1+x^{2} } } \\ {\displaystyle \int \cfrac{dx}{(1+x^{2} )\text{arctan}\,\, x} = \displaystyle \int \cfrac{du}{u} =\ln \left|u\right| +C=\ln \left|\text{arctan}\,\, x\right|+C}$

Example 4. Calculate the indefinite integral of

$\displaystyle \int \cfrac{dx}{x(\ln x)(\ln (\ln x))}$

Solution 4. Use repeated substitutions if necessary to evaluate integrals. Now, which substitution to pick, so many choices? We know the derivative of ln x is 1/x and we see that piece in the integrand so let us try that substitution and see what happens.

$u=\ln x\qquad du=\cfrac{dx}{x}$

$\displaystyle \int \cfrac{dx}{x(\ln x)(\ln (\ln x))} = \displaystyle \int \cfrac{du}{u\ln u}$

We have simplified the integral a bit. Apparently this substitution chops down the chain of iterated logarithms in the denominator one level. Let us chop some more and do another z substitution.

${z=\ln u\quad \, dz=\cfrac{du}{u} }$

$\int \cfrac{du}{u\ln u} = \displaystyle \int \cfrac{dz}{z} =\ln \left|z\right| +C=\ln \left|\ln u\right|+C=\ln \left|\ln (\ln x)\right|+C$

Example 5. Calculate the definite integral of

$\displaystyle \int _{0}^{\pi }x\text{cos}\,\, (x^{2} )dx$

Solution 5. We notice a function and its derivative in the integrand, namely x^2 and 2x. We can write

$u = x^{2} \quad du=2xdx$

$\displaystyle \int _{0}^{\pi }x\text{cos}\,\, (x^{2} )dx =\cfrac{1}{2} \displaystyle \int _{0}^{\pi ^{2} }\text{cos}\,\, udu=\cfrac{1}{2} \left. \text{sin}\,\, u\right| _{u=0}^{u=\pi ^{2} } =\cfrac{1}{2} \text{sin}\,\, (\pi ^{2} )$

Notice that u=x^2 so when x=0 then u=0 and when x= pi then u=pi^2. This demonstrates how to update the limits of integration. Definite integrals requires no back substitution as long as the limits of integration are updated along the way. Updating the limits may save some calculation time.

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