8.5. TRIG. SUBSTITUTIONS 199

partition becomes increasingly small so that the rectangles get increasingly thin thatwhat occurs in the limit should be the definition of the area between the two graphs.But this limit is defined as

∫ ba | f (x)−g(x)|dx. Find the area between the two given

graphs on the given interval.

(a) f (x) = x,g(x) = x− x3,x ∈ [0,3](b) f (x) = sin(x) ,g(x) = cos(x) ,x ∈ [0,2π]

(c) f (x) = ex,g(x) = ln(x) ,x ∈ [1,2]

8.5 Trig. SubstitutionsCertain antiderivatives are easily obtained by making an auspicious substitution involvinga trig. function. The technique will be illustrated by presenting examples.

Example 8.5.1 Find∫ 1

(x2+2x+2)2 dx.

Complete the square as before and write∫ 1

(x2 +2x+2)2 dx =∫ 1(

(x+1)2 +1)2 dx

Use the following substitution next.

x+1 = tanu (8.4)

so dx =(sec2 u

)du. Therefore, this last indefinite integral becomes∫ sec2 u

(tan2 u+1)2 du =∫ (

cos2 u)

du =∫ 1+ cos2u

2du

=u2+

sin2u4

+C =u2+

2sinucosu4

+C

Next write this in terms of x using the following device based on the following picture.

x+1

1

√(x+1)2 +1

u

In this picture which is descriptive of 8.4, sinu = x+1√(x+1)2+1

and cosu = 1√(x+1)2+1

.

Therefore, putting in this information to change back to the x variable,∫ 1

(x2 +2x+2)2 dx

=12

arctan(x+1)+12

x+1√(x+1)2 +1

1√(x+1)2 +1

+C

=12

arctan(x+1)+12

x+1

(x+1)2 +1+C.