Integration by Substitution
"Integration by Substitution" (also called "u-Substitution" or "The Reverse Chain Rule") is a method to find an integral , but only when it can be set up in a special way.
The first and most vital step is to be able to write our integral in this form:
Like in this example:
When our integral is set up like that, we can do this substitution :
Then we can integrate f(u) , and finish by putting g(x) back as u .
Example: ∫ cos(x 2 ) 2x dx
We know (from above) that it is in the right form to do the substitution:
Now integrate:
∫ cos(u) du = sin(u) + C
And finally put u=x 2 back again:
sin(x 2 ) + C
So ∫ cos(x 2 ) 2x dx = sin(x 2 ) + C
That worked out really nicely! (Well, I knew it would.)
Let's just run through that again in a step-by-step manner:
But this method only works on some integrals of course, and it may need rearranging:
Example: ∫ cos(x 2 ) 6x dx
Oh no! It is 6x , not 2x like before. Our perfect setup is gone.
Never fear! Just rearrange the integral like this:
∫ cos(x 2 ) 6x dx = 3 ∫ cos(x 2 ) 2x dx
(We can pull constant multipliers outside the integration, see Rules of Integration .)
Then go ahead as before:
3 ∫ cos(u) du = 3 sin(u) + C
Now put u=x 2 back again:
3 sin(x 2 ) + C
Now let's try a slightly harder example:
Example: ∫ x/(x 2 +1) dx
Let me see ... the derivative of x 2 +1 is 2x ... so how about we rearrange it like this:
∫ x/(x 2 +1) dx = ½ ∫ 2x/(x 2 +1) dx
Then we have:
Then integrate:
½ ∫ 1/u du = ½ ln | u | + C
Now put u=x 2 +1 back again:
½ ln(x 2 +1) + C
And how about this one:
Example: ∫ (x+1) 3 dx
Let me see ... the derivative of x+1 is ... well it is simply 1.
So we can have this:
∫ (x+1) 3 dx = ∫ (x+1) 3 · 1 dx
∫ u 3 du = u 4 4 + C
Now put u=x+1 back again:
(x+1) 4 4 + C
We can take that idea further like this:
Example: ∫ (5x+2) 7 dx
If it was in THIS form we could do it:
∫ (5x+2) 7 5 dx
So let's make it so by doing this:
1 5 ∫ (5x+2) 7 5 dx
The 1 5 and 5 cancel out so all is fine.
And now we can have u=5x+2
And then integrate:
1 5 ∫ u 7 du = 1 5 u 8 8 + C
Now put u=5x+2 back again, and simplify:
(5x+2) 8 40 + C
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Example 4.1.11: Integration by alternate methods. Evaluate ∫ x2 + 2x + 3 √x dx with, and without, substitution. Solution. We already know how to integrate this particular example. Rewrite √x as x1 2 and simplify the fraction: x2 + 2x + 3 x1 / 2 = x3 2 + 2x1 2 + 3x − 1 2. We can now integrate using the Power Rule:
4.Evaluate the following inde nite integral. This can be done with only one substitution, but may be easier to approach with two. Hint: Use u= x2 for the rst substitution, rewrite the integral in terms of u, and then nd a substitution v= f(u). Z xsin(x 2)cos8(x)dx Let v= cos(u) = cos(x2). Then dv= 2xsin(x2)dx, so Z xsin(x2)cos8(x2)dx = 1 2 Z v8 ...
When our integral is set up like that, we can do this substitution: Then we can integrate f (u), and finish by putting g (x) back as u. Like this: Example: ∫ cos (x 2) 2x dx. We know (from above) that it is in the right form to do the substitution: Now integrate: ∫ cos (u) du = sin (u) + C.
In this section we examine a technique, called integration by substitution, to help us find antiderivatives. Specifically, this method helps us find antiderivatives when the integrand is the result of a Chain Rule derivative. At first, the approach to the substitution procedure may not appear very obvious.
6 for integrating powers of a variable tells you to increase the power by 1 and then divide by the new power. In the integral given by Equation (1) there is still a power 5, but the integrand is more compli-cated due to the presence of the term x + 4. To tackle this problem we make a substitution. We let u = x + 4.
22. Use an appropriate trigonometric identity followed by a reasonable substitution to evaluate Z tanxdx 23. It can be shown that 32x2 +77x+49 (3x+1)(4x+5)2 = 2 3x+1 1 (4x+5)2. Use this fact to evaluate Z 32x2 +77x+49 (3x+1)(4x+5)2 dx. 24. Using the substitution x= sin for ˇ 2 ˇ 2, evaluate Z p 1 x2 dx. Express your answer completely in terms ...
Example 4.3.1. Determine the general antiderivative of. h(x) = (5x − 3)6. Check the result by differentiating. For this composite function, the outer function f is f(u) = u6, while the inner function is u(x) = 5x − 3. Since the antiderivative of f is F(u) = 1 7u7 + C, we see that the antiderivative of h is.
4.5 Integration by Substitution. Homework Part 2. Homework Part 1. Calculus Home Page. Class Notes: Prof. G. Battaly, Westchester Community College, NY. of composite functions. All derivatives here use the Chain Rule to find the derivative. 2 3 (x ) sec 2. dy/dx = 3x.
24.Using the substitution x= sin for ˇ 2 ˇ 2, evaluate Z p 1 x2 dx. Express your answer completely in terms of the variable x. HINT - The following trigonometric identities will be helpful: sin 2 + cos = 1, cos2 = 1 2 (1+cos(2 ), and sin(2 ) = 2sin cos 1 2 x p 1 x2 + 1 2 sin 1 x+C 4
he numerator. In Example 3 we had 1, so the de. ree was zero. To make a successful substitution, we would need u to be a degree 1 polynomia. (0 + 1 = 1). Obviously the polynomial on the denominator. was degree 2. So we forced a degree 1 polynomial to appear by completing the. . tice ProblemsTry some of the p. oblems below. If you get stuck,