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.
dx 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
((sin(2x))3)
---6---
^{′} = cos
(2x )
sin^{2}
(2x)
which verifies the answer is right.
Here is another example.
Example 7.1.5Find∫
√3------
2x+ 7
xdx.
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
u^{7∕3}−
21
16
u^{4∕3} + C
=
-3
28
(2x + 7)
^{7∕3}−
21
16
(2x+ 7)
^{4∕3} + C
Example 7.1.6Find∫x3^{x2
}dx
Let u = 3^{x2
} so that
ddux
= 2xln
(3)
3^{x2
} and
2dlun(3)
= 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
--1--
2ln(3)
3^{x2
} + C.
Example 7.1.7Find∫
cos^{2}
(x)
dx
Recall that cos
(2x)
= cos^{2}
(x)
− sin^{2}
(x)
and 1 = cos^{2}
(x )
+ sin^{2}
(x)
. Then
subtracting and solving for cos^{2}
(x)
,
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.
Example 7.1.8Find∫
tan
(x )
dx
Let u = cosx so that du = −sin
(x)
dx. Then writing the antiderivative in terms of u,
this becomes ∫
−1
u
du. At this point, recall that
(ln |u|)
^{′} = 1∕u. Thus this antiderivative is
−ln
|u|
+ C = ln
|| −1||
u
+ C and so ∫
tan
(x)
dx = ln
|secx|
+ C.
This illustrates a general procedure.
Procedure 7.1.9∫
f′(x)
f(x)
dx = ln
|f (x)|
+ C.
This follows from the chain rule and the derivative of x → ln