Calculus/Integration techniques/Recognizing Derivatives and the Substitution Rule

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Integration techniques/Recognizing Derivatives and the Substitution Rule

After learning a simple list of antiderivatives, it is time to move on to more complex integrands, which are not at first readily integrable. In these first steps, we notice certain special case integrands which can be easily integrated in a few steps.

Recognizing Derivatives and Reversing Derivative RulesEdit

If we recognize a function g(x) as being the derivative of a function f(x), then we can easily express the antiderivative of g(x):

\int g(x)\, dx = f(x) + C.

For example, since

\frac{d}{dx} \sin x = \cos x

we can conclude that

\int \cos x\, dx = \sin x + C.

Similarly, since we know e^x is its own derivative,

\int e^x \, dx = e^x + C.


The power rule for derivatives can be reversed to give us a way to handle integrals of powers of x. Since

\frac{d}{dx} x^n = n x^{n-1},

we can conclude that

\int n x^{n-1} \, dx = x^n + C,

or, a little more usefully,

\int x^n \, dx = \frac{x^{n+1}}{n+1} + C.

Integration by SubstitutionEdit

For many integrals, a substitution can be used to transform the integrand and make possible the finding of an antiderivative. There are a variety of such substitutions, each depending on the form of the integrand.

The objective of Integration by substitution is to substitute the integrand from an expression with variable x to an expression with variable  u where u \,=\, g(x)

Theory

We want to transform the Integral from a function of x to a function of u

\int_{x=a}^{x=b}f(x)\, dx\,\rightarrow\,\int_{u=c}^{u=d}h(u)\, du

Starting with

u\,=\,g(x)


Steps

\int_{x=a}^{x=b}f(x)\, dx\,  =\int_{x=a}^{x=b}f(x)\,{\operatorname{d}\!u\over\operatorname{d}\!u}\, dx\,      (1) ie   {\operatorname{d}\!u\over\operatorname{d}\!u}\,=\,1
=\int_{x=a}^{x=b}(f(x)\,{\operatorname{d}\!x\over\operatorname{d}\!u})({\operatorname{d}\!u\over\operatorname{d}\!x})\, dx\,      (2) ie   {\operatorname{d}\!x\over\operatorname{d}\!u}{\operatorname{d}\!u\over\operatorname{d}\!x}\,=\,{\operatorname{d}\!u\over\operatorname{d}\!u}\,=\,1
=\int_{x=a}^{x=b}(f(x)\,{\operatorname{d}\!x\over\operatorname{d}\!u})g'(x)\, dx\,      (3) ie   {\operatorname{d}\!u\over\operatorname{d}\!x}\,=\,g'(x)
=\int_{x=a}^{x=b}h(g(x))g'(x)\, dx\,      (4) ie   Now equate (f(x)\,{\operatorname{d}\!x\over\operatorname{d}\!u}) with h(g(x))
=\int_{x=a}^{x=b}h(u)g'(x)\, dx\,      (5) ie   g(x)\,=\,u
=\int_{u=g(a)}^{u=g(b)}h(u)\, du\,      (6) ie    du\,=\,{\operatorname{d}\!u\over\operatorname{d}\!x} dx\,=\,g'(x)\, dx\,
=\int_{u=c}^{u=d}h(u)\, du\,      (7) ie   We have achieved our desired result

Procedure

  • Calculate g'(x)\,=\,{\operatorname{d}\!u\over\operatorname{d}\!x}
  • Calculate h(u) which is f(x)\,{\operatorname{d}\!x\over\operatorname{d}\!u}\,=\,\frac{f(x)}{g'(x)} and make sure you express the result in terms of the variable u
  • Calculate c\,=\,g(a)
  • Calculate d\,=\,g(b)

Integrating with the derivative presentEdit

If a component of the integrand can be viewed as the derivative of another component of the integrand, a substitution can be made to simplify the integrand.

For example, in the integral

\int 3x^2 (x^3+1)^5 \, dx

we see that 3x^2 is the derivative of x^3+1. Letting

u=x^3+1

we have

\frac{du}{dx} = 3x^2

or, in order to apply it to the integral,

du = 3x^2 dx.

With this we may write \int 3x^2 (x^3+1)^5 \, dx = \int u^5 \, du = \frac{1}{6} u^6 + C = \frac{1}{6} (x^3+1)^6 + C.

Note that it was not necessary that we had exactly the derivative of u in our integrand. It would have been sufficient to have any constant multiple of the derivative.

For instance, to treat the integral

\int x^4 \sin(x^5) \, dx

we may let u=x^5. Then

du = 5x^4 \, dx

and so

\frac{1}{5} du = x^4 \, dx

the right-hand side of which is a factor of our integrand. Thus,

\int x^4 \sin(x^5) \, dx = \int \frac{1}{5} \sin u \, du = -\frac{1}{5} \cos u + C  = -\frac{1}{5} \cos x^5 + C.

In general, the integral of a power of a function times that function's derivative may be integrated in this way. Since \frac{d[g(x)]}{dx}=g'(x),

we have dx=\frac{d[g(x)]}{g'(x)}.

Therefore, \int g'(x)[g(x)]^n dx\, = \int g'(x)[g(x)]^n \frac{d[g(x)]}{g'(x)}
=\int [g(x)]^n d[g(x)]
=\frac{[g(x)]^{n+1}}{n+1}


There is a similar rule for definite integrals, but we have to change the endpoints.

Substitution rule for definite integrals

Assume u is differentiable with continuous derivative and that f is continuous on the range of u. Suppose c=u(a) and d=u(b). Then \int_a^b f(u(x)) \frac{du}{dx} dx = \int_c^d f(u) du.

ExamplesEdit

Consider the integral


\int_{0}^2 x \cos(x^2+1) \,dx

By using the substitution u = x2 + 1, we obtain du = 2x dx and

\int_{0}^2 x \cos(x^2+1) \,dx = \frac{1}{2} \int_{0}^2 \cos(x^2+1) 2x \,dx
= \frac{1}{2} \int_{1}^{5}\cos(u)\,du
= \frac{1}{2}(\sin(5)-\sin(1)).

Note how the lower limit x = 0 was transformed into u = 02 + 1 = 1 and the upper limit x = 2 into u = 22 + 1 = 5.

Proof of the substitution ruleEdit

We will now prove the substitution rule for definite integrals. Let H be an anti derivative of h so

 H'(u) = h(u).

Suppose we have a differentiable function, g(x) such that u\,=\,g(x), and numbers c=g(a) and d=g(b) derived from some given numbers, a and b.
By the Fundamental Theorem of Calculus, we have

 \int_c^d h(u) du = H(d) - H(c).

Next we define a function F by the rule

 F(x) = H(g(x)) = H(u)\,.

Naturally

f(x)\,=\,F'(x)\,=\,\frac{dF}{dx}

Then by the Chain rule F is differentiable with derivative

 \frac{dF}{dx} = \frac{dH}{du}\frac{du}{dx}= h(u) \frac{du}{dx} = h(g(x)) \frac{du}{dx}\,.

Integrating both sides with respect to x and using the Fundamental Theorem of Calculus we get

 \int_a^b h(g(x)) \frac{du}{dx}dx =\int_a^b \frac{dF}{dx} dx = F(b) - F(a).

But by the definition of F this equals

 F(b) - F(a) =  H(g(b)) - H(g(a)) = H(d) - H(c) = \int_c^d h(u) du.

Hence

\int_a^b f(x) dx\,=\,\int_a^b h(g(x)) \frac{du}{dx}dx = \int_c^d h(u) du.

which is the substitution rule for definite integrals.

ExercisesEdit

Evaluate the following using a suitable substitution.

1. \int\frac{19}{\sqrt{9x-38}}dx

\frac{38\sqrt{9x-38}}{9}+C

2. \int-15\sqrt{9x+43}dx

-\frac{10(9x+43)^{3/2}}{9}+C

3. \int\frac{17\sin(x)}{\cos(x)}dx

-17\ln|\cos(x)|+C

4. \int5\cos(x)\sin(x)dx

\frac{5\sin^{2}(x)}{2}+C

5. \int_{0}^{1}-\frac{10}{(-5x-32)^{4}}dx

-\frac{17885}{2489696256}

6. \int-3e^{3x+12}dx

-e^{3x+12}+C

Solutions

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Integration techniques/Recognizing Derivatives and the Substitution Rule
Last modified on 3 January 2014, at 06:25