The Calculus Primer (2011)
Part XIII. Special Methods of Integration
Chapter 49. INTEGRATION BY SUBSTITUTION; CHANGE OF VARIABLE
13—5. Algebraic Substitution. Frequently an expression to be integrated can be transformed, by the suitable substitution of a new variable, into one of the fundamental standard forms. Some of the simpler kinds of such substitutions will now be illustrated.
EXAMPLE 1. Find
Solution.
Let
Substituting (3x − 2)½ for z:
EXAMPLE 2. Find
Solution.
Let z = (3x + l)⅓;
then dx = z2 dz, and (3x + 1)⅔ = z2.
Substituting (3x + 1)⅓ = z:
EXAMPLE 3. Find
Solution. Let x = z; then x = z6, dx = 6z5 dz, x⅓ = z3, and x⅓ = z2.
Substituting for z, z2, and z3:
EXAMPLE 4. Find
Solution. Let z = ; then x = z2 + 3, and dx = 2z dz.
Substituting
EXERCISE 13—3
Find the following:
Verify the following:
13—6. Trigonometric Substitutions. When the integrand contains expressions such as the integration may be performed by using the following trigonometric substitutions:
I.For we put v = a sin θ; the expression then becomes:
II.For we put v = a tan θ; the expression then becomes:
III.For we put v = a sec θ; the expression then becomes:
EXAMPLE 1. Find
Solution. Let a2 = 9, a = 3; x = 3 sin z, dx = 3 cos z dz; Therefore,
But, x = 3 sin z; from the triangle of reference we have z = arc sin , and cos
hence
EXAMPLE 2. Find
Solution. Let x = 2 tan z; dx = 2 sec2 z dz;
= log (sec z + tan z).
Therefore, from the triangle of reference,
EXAMPLE 3. Find
Solution. Let x = a sec z; dx = a sec z tan z dz; = a tan z.
Therefore,
But sec z = ; hence, from the triangle:
EXERCISE 13—4
Verify the following: