2. Two Methods for Making
Substitutions in Definite
Integrals
• “Conveniently” ignore the limits of
integration and revisit them at the end
of the process.
• Recalculate the limits in the very
beginning so you don’t have to revisit
them at the end.
3. Let’s see how does that
works without
recalculating limits of
integration.
2
∫ x( x
0
3
2
1/2
I
+1) dx =
∫(x
2
+1)
2
3 I
1
xdx =
2
4 2
∫ u du =
3
( x 2 +1) 1 2
1u
= ( 2 +1) 4 − ( 0 2 +1) 4 =
du = 2xdx
=
8
2 4
8
0
1
624
= 78
[ 625 −1] =
8
8
u = x 2 +1
4
4. How about
recalculating limits of
integration?
π
8
∫ sin
5
2x cos2xdx =
π
1/2 8
I
0
u = sin 2x
du = 2 cos2xdx
x = 0 → u = sin(0) = 0
π 1
π
x = → u = sin ÷=
4
8
2
∫ sin 2x
5
0
1
2
1u
2 6 0
6
2
I
1
cos2xdx =
2
1
2
∫
u5du =
0
1 6
1
6
=
÷ − ( 0) =
12 2
1 1 1
− 0 ÷=
12 8 96
5. Practice Time!!!
9
5
1. ∫ ( 2x − 5) ( x − 3) dx =
2
3
4
dx
2. ∫
=
0 1− x
ln3
3. ∫ e x ( 1+ e
0
2
1
x 2
)
dx =
dx
4. ∫ 2
=
1 x − 6x + 9
2
5. ∫ xe
1
− x2
=
6. More!!!
π
6
1. ∫ 2 cos3xdx =
0
(
ln 2
2.
∫
ln 2
3
)
e− x dx
1− e
−2 x
=
2
3. ∫ 5x −1dx =
1
ln 5
4. ∫ e x ( 3 − 4e x ) dx =
0
π
6
5. ∫ tan 2θ dθ =
0
7. More!!!
π
6
1. ∫ 2 cos3xdx =
0
(
ln 2
2.
∫
ln 2
3
)
e− x dx
1− e
−2 x
=
2
3. ∫ 5x −1dx =
1
ln 5
4. ∫ e x ( 3 − 4e x ) dx =
0
π
6
5. ∫ tan 2θ dθ =
0
