Unizor - Indefinite Integrals - Variable Substitution

Notes to a video lecture on http://www.unizor.com

Indefinite Integral -
Variable Substitution

Assume, you know that
∫ f(x) dx = g(x) + C

What immediately follows from this is that the derivative of g(x) is the original function f(x):
g^I(x) = f(x)
or, equivalently,
dg(x) = f(x) dx

Given that, consider a derivative of a compound function g(w(x)):
[g(w(x))]^I = f(w(x))·w^I(x)
or, equivalently,
dg(w(x)) = f(w(x))·w^I(x) dx = f(w(x)) dw(x)

From equality of derivatives or differentials of two functions
f(w(x)) dw(x) = dg(w(x))
follows that original functions (before the differentiation or obtained from the differential by integration) differ only by a constant.
Therefore,
∫ f(w(x)) dw(x) = ∫ dg(w(x))

Since diiferentiation and integration are inversed to each other, the right part represents a function g(w(x)) (plus constant, as usually with integration).
Therefore,
∫ f(w(x)) dw(x) = g(w(x)) + C
or, equivalently,
∫ f(w(x))·w^I(x) dx = g(w(x)) + C

Compare this with original relationship between f(x) and g(x) above:
∫ f(x) dx = g(x) + C
As we see, to integrate f(w(x))·w^I(x), it is sufficient to integrate f(x) and substitute w(x) instead of x in the answer.

This is a substitution rule of integration.

Example 1:

∫ x·e^(x²) dx
To find this integral, notice that we can do the following:
f(x) = e^x
∫ f(x) dx = e^x + C
w(x) = x²
w^I(x) = 2x
f(w(x)) = e^(x²)
f(w(x))·w^I(x) = e^(x²)·2x = e^(x²)·(x²)^I
f(w(x)) dw(x) = e^(x²) d(x²)

Using the above substitution, we obtain the answer:
∫ x·e^(x²) dx = (1/2)∫ e^(x²) d(x²) = (1/2)e^(x²) + C

Checking by differentiating the answer (D_x is the operation of differentiation):
D_x (1/2)e^(x²) = (1/2)e^(x²)·D_x x² = (1/2)e^(x²)·2x = x·e^(x²)
The end

Example 2:

∫ sin(x)·cos²(x) dx
To find this integral, notice that we can do the following:
f(x) = x²
∫ f(x) dx = x³/3 + C
w(x) = cos(x)
w^I(x) = −sin(x)
f(w(x)) = cos²(x)
f(w(x))·w^I(x) = −cos²(x)·sin(x) = cos²(x)·(cos(x))^I
f(w(x)) dw(x) = −cos²(x) dcos(x)

Using the above substitution, we obtain the answer:
∫ sin(x)·cos²(x) dx = −∫ cos²(x) dcos(x) = −cos³(x)/3 + C

Checking by differentiating the answer:
D_x −cos³(x)/3 = −3cos²(x)·(−sin(x))/3 = cos²(x)·sin(x)
The end.

Example 3:

∫ ln(sin(x))·cos(x) dx
To find this integral, notice that
(sin(x))^I = cos(x),
we use sin(x) as an inner function:
ln(sin(x))·cos(x) = ln(sin(x))·(sin(x))^I
Also recall from the previous lecture that
∫ ln(x) dx = x·ln(x) − x + C

Using the above substitution, we obtain the answer:
∫ ln(sin(x))·cos(x) dx = ∫ ln(sin(x)) dsin(x) = sin(x)·ln(sin(x)) − sin(x) + C

Checking by differentiating the answer:
D_x (sin·ln(sin(x)) − sin(x)) = ln(sin(x))·cos(x) + sin(x)·(1/sin(x))·cos(x) − cos(x) = ln(sin(x))·cos(x)
The end.