CALCULUS II

TUTORIAL 2 ALL COHORTS 2 SEPTEMBER 2022

TECHNIQUES OF INTEGRATION

1. THE FUNDAMENTAL THEOREM OF CALCULUS/POWER RULE

The Fundamental Theorem of Calculus tells us that the exact area under a continuous function f
over an interval [ a , b ]is calculated directly using a definite integral:

∫ f ( x ) dx=F ( b )−F (a)

a

The function is any antiderivative of (we usually set the constant of integration equal to 0).

1 n+1
Power Rule of Antidifferentiation: ∫ x dx=
n
x + c, n ≠ 1
n+1

Integrate the following:

1. i. ii. iii.

∫ ( 5 x +7 ) dx
2. i. ii. iii.

3. i. ∫ ( x +1 )2 dx ii. ∫ ( t2 +5 t +1 ) dt iii. ∫ ( x +1 )2 dx

( 1
)
1
4. i ∫ x+
2

x
2
dx ii. ∫ ( t3 +t 2 ) dt iii. ∫t 2
dt

5. i. ∫ 3 √ w dw ii. ∫ (x 3¿ +5 x 2 +6) dx ¿ iii. ∫ ( 3 2

)
− dt
t t2

2 5 1
1
6. i. .∫ 2
dt ii. ∫ ( x 3−π x 2 ) dx iii. ∫ ( y 2+ y 4 ) dy
1 t 2 0

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 1 9 1

7. i. ∫ 2 e dtt
ii. ∫ √ x dx iii. ∫ ( 6 q2 + 4 ) dq
0 4 0

5 4 2

8. i. ∫ 3 x dx2
ii. ∫ √1x dt iii. ∫ ( 3t 2 +4 t +3 ) dt
0 1 0

∫ ( 1y + y +1 ) dy
2 2

9. i. ii. ∫ 1x dx iii. ∫ x12 dx

1 1

10. i. ∫ ( 4 t + 7 ) dt ii. ∫ ( x +1 )2 dx iii. ∫ ( x + √1x ) dx

2. INTEGRATION BY SUBSTITUTION

The following formulas provide a basis for an integration technique called substitution:

ur +1
A. ∫ u r du= r +1
+ c , assuming r ≠−1

B. ∫ eu du=eu + c
1 1
C. ∫ u du=ln|u|+c , or ∫ u du=ln ⌈ u ⌉ +c , u>0

Other identities:

−1
a . ∫ sin ( ax +b ) dx= cos ( ax+ b ) +c
a

1
b .∫ cos ( ax +b ) dx = sin ( ax +b ) + c
a

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 1
∫ ax +b dx=ln|ax+ b|+c

Integrate the following using Substitution:

1
1. i. ∫ ( x +4 ) dx ii.∫ cos ( 3 x + 4 ) dx
5
iii ∫ 1−2 x dx
3 1

i. ∫ ( 9+ x ) dx ii.∫ ( x−2 ) dx iii. ∫ ( x +5 ) dx

2 3 4
2.
1 0

1
i. ∫ ( 2 x−1 ) dx ii.∫ ( 1−x ) dx iii. ∫ sin ( 7 x−3 ) dx
7 3
3.
−1

1
i. ∫ e ii.∫ iii. ∫ 2 x √1+ x dx
3 x−2 2
4. dx dx
7 x+ 5

4x sin √ x
5. i. ∫ dx ii. ∫ dx iii. ∫ 5 x √1−x2 dx
√2x 2
+1 √x

dx x
3
6. i. ∫ ii. ∫ x ( 1+ x ) dx 4 5 3
iii. ∫ 4 dx
√ x +16
2
√ x ( 1+ √ x )
1
cos x x
3
7. i. ∫ dx ii. ∫ dx iii. ∫ 5 x 2 √ 1−x 3 dx
( 5+sin x )2 0 √x 4
+12

8. i. ∫ e
cos x
sin x dx ii. ∫ e
sinx
cos x dx iii. ∫ x
3
√ x 4 +1 dx
0

1 1 1
5
i. ∫ x ( x +1 ) dx ii. . ∫ x √ x +1dx iii. ∫
2 5 2
9. dx
0 0 0 ( 5 x+1 )2

2 3 1
x +1 dx 5
10. i. ∫ 2
dx ii. ∫ iii. ∫( 2
dx
1 (x +2x)
2
0 x+ 1 0 5 x+1 )

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3. INTEGRATION BY PARTS

Recall the Product Rule for differentiation,

d dv du
( uv )=u + v
dx dx dx

Integrating both sides with respect to x , we get

dv du
uv=∫ u dx +∫ v dx
dx dx

¿ ∫ udv +∫ vdu

Solving for ∫ udv , we get the following theorem,

∫ udv=uv−∫ vdu
Use Integration by Parts to integrate the given functions

1. i. ii. iii.

2. i. ii. iii.

3. i. ii. iii.

4. i. ii. iii.

5. i. ii. iii.

6. i. ii. iii.

7. i. ii. iii.

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 3

8. i. ∫ ln ( x +7 ) dx ii. ∫ ln ( x + 4 ) dx iii. ∫ x 2 ln ( 5 x ) dx
0

9. i. ∫ x √ x +2 dx ii. ∫ ( x−1 ) ln x dx iii. ∫ ln ( x )2 dx

10. i. ∫ x ln √ x dx ii. ∫ x e−x dx iii. ∫ x 2 ( 2 x ) dx

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