Part II: The Answers
508
Answers
801–900
847.
4 1 2
3
2
+
+
+
x
x
C
Here’s the given integral:
4 12
1 2
3
2
+
+
+
∫
x
x
x
dx
If you let u = 1 + 2x + 3x
2
, then you get du = (2 + 6x)dx, or 2du = (4 + 12x)dx. Substitute
this value into the original integral:
2
2
2 1 2
4 1 2
3
1 2
1 2
2
u
du
u du
u
C
x
x
C
∫
∫
=
=
+
=
+
+
+
−
848.
−
+
3
4
4 3
cot x C
Here’s the given integral:
cot csc
x
xdx
3
2
∫
If you let u = cot x, then you get du = –csc
2 
x
 
dx, or –du = csc
2 
x
 
dx. Substitute this value
into the original integral:
−
=−
= −
+
= −
+
= −
+
∫
∫
u du
u du
u
C
x
C
x C
3
1 3
4 3
4 3
4 3
4 3
3
4
3
4
(cot )
cot
849.
− +
2 2
4
Here’s the given integral:
x
x dx
sin
2
0
2
( )
∫
π
If you let u = x
2
, then du = 2x
 
dx, or 1
2
du x dx
=
. You can also find the new limits of
integration by using the original limits of integration along with the u-substitution.
When x = π
2
, u =





 =
π
π
2
4
2
. Likewise, if x = 0, then u = (0)
2
= 0.