INTEGRATION BY SUBSTITUTION; CHANGE OF VARIABLE - Special Methods of Integration - The Calculus Primer

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 image image

Solution.

Let image

Substituting (3x − 2)½ for z:

EXAMPLE 2. Find image

Solution.

Let z = (3x + l);

then image dx = z2 dz, and (3x + 1) = z2.

image

Substituting (3x + 1) = z:

EXAMPLE 3. Find image image

Solution. Let ximage = z; then x = z6, dx = 6z5 dz, x = z3, and x = z2.

Substituting for z, z2, and z3:

EXAMPLE 4. Find image

Solution. Let z = image; then x = z2 + 3, and dx = 2z dz.

Substituting image

EXERCISE 13—3

Find the following:

Verify the following:

13—6. Trigonometric Substitutions. When the integrand contains expressions such as image the integration may be performed by using the following trigonometric substitutions:

I.For image we put v = a sin θ; the expression then becomes: image

II.For image we put v = a tan θ; the expression then becomes: image

III.For imagewe put v = a sec θ; the expression then becomes: image

EXAMPLE 1. Find image

Solution. Let a2 = 9, a = 3; x = 3 sin z, dx = 3 cos z dz; image Therefore,

But, x = 3 sin z; from the triangle of reference we have z = arc sin image, and cos image hence

EXAMPLE 2. Find image image

Solution. Let x = 2 tan z; dx = 2 sec2 z dz; image

image

= log (sec z + tan z).

Therefore, from the triangle of reference,

EXAMPLE 3. Find image

Solution. Let x = a sec z; dx = a sec z tan z dz; image = a tan z.

Therefore,

But sec z = image; hence, from the triangle:

EXERCISE 13—4

Verify the following: