integration by substitution

Recall that the chain rule says

[ f(g(x)) ]' = f '(g(x)) g'(x)

So if we antidifferentiate this we get

ò [ f(g(x)) ]' dx	=	ò f '(g(x)) g'(x) dx, so
f(g(x)) + C	=	ò f '(g(x)) g'(x) dx

(because on the left-hand-side we're just taking the antiderivative of a derivative). What does this mean for us? When we have an integrand that is a product, if it matches the pattern on the right-hand side of the equation above, that is, if one term of the product is the derivative of the part of another, we can use substitution. Next we show how this works without substitution, and after that we use substitution to do it much more easily.

Example: Let's look at

ò x² cos(x³ + 4) dx.

There is a composition here, cos(x³ + 4), so g(x) = x³ + 4. Then g'(x) = 3 x², so g'(x)/ 3 = x², and our integral is actually

ò (g'(x)/ 3) cos(g(x)) dx = (1/ 3) ò cos(g(x)) g'(x) dx (factoring out constants).

This is hopeful, because it says if cos(x) is f '(x) for some f, we have exactly the reversed chain rule problem that we looked at above. And, of course, it is:

if f(x) = sin(x), then f '(x) = cos(x).

So our integral is "easy":

ò x² cos(x³ + 4) dx	=	(1/ 3) ò cos(g(x)) g'(x) dx
	=	(1/3) ò f '(g(x)) g'(x) dx
	=	(1/ 3) f(g(x)) + C
	=	(1/ 3) sin(x³ + 4) + C.

Substitution Explanation:

Substitution makes it easier to see the composition in an integrand. To use it,

pick w to be the "inner" function (g(x) above),
find dw/dx and solve for dw,
multiply both sides of the equation for dw by any constants you need to make it match terms in the integral,
substitute w and dw into the integral to get rid of all terms involving x,
find an antiderivative, and
plug w(x) into the antiderivative.

Note that if after step 4 you still have terms involving x left in the integrand, either substitution doesn't work, or it's necessary to solve the equation w = g(x) for x and plug that in as well.

Let's do this with our example.

Example, part 2: Find

ò x² cos(x³ + 4) dx.

The inner function is w = x³ + 4,
dw/dx = 3 x², so dw = 3 x² dx,
to make this match the integrand, multiply by 1/3: (1/3) dw = x² dx.
Then the integral becomes

ò (1/3) cos(w) dw,

and all terms in x are in fact gone. Thus
we can antidifferentiate: ò (1/3) cos(w) dw = (1/3) ò cos(w) dw = (1/3) sin(w) + C, and
plugging back in for w, ò x² cos(x³ + 4) dx = (1/3) sin(x³ + 4) + C.

In some cases, we can change the appearance of an integrand by using substitution and therefore make it easier to see how to find the antiderivative. For example, consider the integrals

ò x sqrt(x + 3) dx and ò x^{3/ 2} - 3 x^{1/ 2} dx.

These are the same integral in disguise, but while the first of these is not obvious, the second consists of two terms that are easy to antidifferentiate (see the basic antidifferentiation page).

Picking w to Rewrite the Integrand:

To use substitution to rewrite an integrand, try picking the substitution variable w to be some part of the integrand that is an "inner" function in a composition; or part that is under a root sign; or part that is in the denominator of a fraction; or a combination of terms that appears in multiple places in the integrand. Then continue with steps 2-6 as indicated in the previous section.

Example: Let's do the example above. Find

ò x sqrt(x + 3) dx

The term under the root sign is w = x + 3. So

dw/dx = 1, so that dw = dx.
We have just a dx in the integrand, so we don't have to multiply by anything.
Substituting, we get

ò x sqrt(x + 3) dx = ò x sqrt(w) dw

We have a remaining x, but can turn that into a w by solving for x:

w = x + 3, so w - 3 = x, and
ò x sqrt(w) dw = ò (w - 3) sqrt(w) dw.

We can now antidifferentiate: rewriting sqrt(w) as w to the one-half power, we get

ò (w - 3) sqrt(w) dw	=	ò (w - 3) w^{1/ 2} dw
	=	ò (w^{3/ 2} - 3 w^{1/ 2}) dw
	=	(2/5) w^{5/ 2} - 2 w^{3/ 2} + C

(by basic antidifferentiation).

Finally, plugging back in for w,

ò x sqrt(x + 3) dx = (2/5)(x + 3)^{5/ 2} - 2(x + 3)^{3/ 2} + C

Integration by Substitution