Calculus of One and Several Variables

7.1 The Method of Substitution

The method of substitution is based on the following formula which is merely a restatement of the chain rule.

∫ f (g (x ))g′(x)dx = F (g (x))+ C, (7.1)

(7.1)

where F^′

= f. Here are some examples of the method of substitution.

Note it is of the form given in 7.1 with g

= 2⁻¹ sin² and F = ³. Indeed, F^′ = . This is an application of the chain rule. Therefore,

∫ ( ) ∘ ----−-1--2--- 1- ∘(--------2-) 3 sin (x)cos(x) 3 +2 sin (x)dx = 12 12 + 2sin x + C.

This is a special case of 7.1 when g

= 1 + x² and F = u⁷∕7. Therefore, the answer is

∫ ( 2)6 ( 2)7 1+ x 2xdx = 1 + x ∕7+ C.

This equals

1∫ ( )6 1(1+ x2)7 (1 + x2)7 2 1 + x2 2xdx = 2---7----+ C = ---14----+ C.

Actually, it is not necessary to recall 7.1 and massage things to get them in that form. There is a trick based on the Leibnitz notation for the derivative which is very useful and illustrated in the following example.

Let u = sin

. Then = 2cos. Now formally

du-= cos(2x)dx. 2

Thus

∫ ∫ 2 1 2 u3 (sin-(2x))3 cos(2x )sin (2x)dx = 2 udu = 6 + C = 6 + C

The expression

du replaced the expression cosdx which occurs in the original problem and the resulting problem in terms of u was much easier. This was solved and finally the original variable was replaced. When using this method, it is a good idea to check your answer to be sure you have not made a mistake. Thus in this example, the chain rule implies ^′ = cossin² which verifies the answer is right. Here is another example.

In this example u = 2x + 7 so that du = 2dx. Then

∫ √32x-+-7- xdx	= ∫ √3u-- x ◜-◞◟-◝ u-−-7 2 dx ◜◞◟◝ 1du 2 = ∫ ( ) 1u4∕3 − 7u1∕3 4 4 du
	= -3 28 u7∕3 − 21 16 u4∕3 + C
	= -3 28 (2x + 7) 7∕3 − 21 16 (2x+ 7) 4∕3 + C

Let u = 3^x2 so that

= 2xln3^x2 and = x3^x2 dx. Thus

∫ 2 1 ∫ 1 1 2 ( 1 ) x3x dx = 2ln(3) du = 2ln(3) [u+ C ] = 2ln(3)3x + 2-ln-(3) C

Since the constant is an arbitrary constant, this is written as

3^x2 + C.

Recall that cos

= cos² − sin² and  1 = cos² + sin². Then subtracting and solving for cos²,

2 1+ cos(2x) cos (x) = ----2----.

Therefore,

∫ ∫ 1 +cos(2x) cos2(x)dx = ----------dx 2

Now letting u = 2x, du = 2dx and so

∫ ∫ 1+ cos(u) 1 1 1 cos2(x)dx = ---------du = -u + -sin u+ C = - (2x+ sin (2x ))+ C. 4 4 4 4

Also

∫ sin2(x)dx = − 1cosxsinx + 1x+ C 2 2

which is left as an exercise.

Let u = cosx so that du = −sin

dx. Then writing the antiderivative in terms of u, this becomes ∫ du. At this point, recall that ^′ = 1∕u. Thus this antiderivative is −ln + C = ln + C and so ∫ tandx = ln + C.

This illustrates a general procedure.

This follows from the chain rule and the derivative of x → ln

.

This is usually done by a trick. You write as

∫ sec(x)(sec(x)+ tan(x)) ---------------------dx (sec(x)+ tan(x))

and note that the numerator of the integrand is the derivative of the denominator. Thus

∫ sec(x )dx = ln|sec(x)+ tan(x)|+ C.

This is done like the antiderivatives for the secant.

-d-csc(x) = − csc (x)cot(x) dx

and

cot = −csc². Write the integral as

∫ − −-csc(x)(cot(x)+-csc-(x))dx = − ln|cot(x)+ csc(x)|+ C. (cot(x)+ csc(x))

Let the position of the point be r

. Then by definition of velocity, r^′ = t² + 1 so r = + t + C. Now C must be determined. It is assumed that r = 1. Therefore, C = 1 and so r = + t + 1.

It is given that r^′′

= t + 1,r^′ = 1. Therefore, r^′ = + t + 1. Then r = + + t + 2.