1.

Answer the following questions:

3
(a) Evaluate ∫𝑠𝑖𝑛 𝑥 𝑑𝑥.
2 2
●​ Key Idea: Use the identity 𝑠𝑖𝑛 𝑥 = 1 − 𝑐𝑜𝑠 𝑥 and substitution.
3 2 2
●​ Steps: ∫𝑠𝑖𝑛 𝑥 𝑑𝑥 = ∫𝑠𝑖𝑛 𝑥 · 𝑠𝑖𝑛𝑥 𝑑𝑥 = ∫(1 − 𝑐𝑜𝑠 𝑥)𝑠𝑖𝑛𝑥 𝑑𝑥 Let 𝑢 = 𝑐𝑜𝑠𝑥, so 𝑑𝑢 =− 𝑠𝑖𝑛𝑥 𝑑𝑥.
3
2 2 𝑢
= ∫(1 − 𝑢 )(− 𝑑𝑢) = ∫(𝑢 − 1)𝑑𝑢 = 3
− 𝑢+ 𝐶
3
𝑐𝑜𝑠 𝑥
●​ Answer: 3
− 𝑐𝑜𝑠𝑥 + 𝐶

𝑑𝑥
(b) Find ∫ 2 2
.
𝑥 −𝑎

●​ Key Idea: This is a standard integral formula.

2 2
●​ Answer: 𝑙𝑛|𝑥 + 𝑥 −𝑎 | + 𝐶
𝑎𝑥+𝑏
(c) Find ∫𝑒 𝑑𝑥.
●​ Key Idea: Use substitution.
1 𝑎𝑥+𝑏 𝑢 1 1 𝑢
●​ Steps: Let 𝑢 = 𝑎𝑥 + 𝑏, so 𝑑𝑢 = 𝑎 𝑑𝑥 ⇒ 𝑑𝑥 = 𝑎
𝑑𝑢. ∫𝑒 𝑑𝑥 = ∫𝑒 𝑎
𝑑𝑢 = 𝑎
𝑒 + 𝐶
1 𝑎𝑥+𝑏
●​ Answer: 𝑎
𝑒 + 𝐶

𝑠𝑖𝑛𝑥
(d) Find ∫𝑐𝑜𝑠𝑥𝑒 𝑑𝑥.
●​ Key Idea: Use substitution.
𝑢 𝑢
●​ Steps: Let 𝑢 = 𝑠𝑖𝑛𝑥, so 𝑑𝑢 = 𝑐𝑜𝑠𝑥 𝑑𝑥. ∫𝑒 𝑑𝑢 = 𝑒 + 𝐶
𝑠𝑖𝑛𝑥
●​ Answer: 𝑒 + 𝐶
1
𝑑𝑥
(e) Evaluate ∫ 𝑥 −𝑥 .
0 𝑒 +𝑒

𝑥
●​ Key Idea: Multiply numerator and denominator by 𝑒 and use substitution.
1 𝑥
𝑒 𝑑𝑥 𝑥 𝑥 0 1
●​ Steps: ∫ 2𝑥 Let 𝑢 = 𝑒 , so 𝑑𝑢 = 𝑒 𝑑𝑥. When 𝑥 = 0, 𝑢 = 𝑒 = 1. When 𝑥 = 1, 𝑢 = 𝑒 = 𝑒.
0 𝑒 +1
𝑒
𝑑𝑢 𝑒
∫ 2 = [𝑎𝑟𝑐𝑡𝑎𝑛𝑢]1 = 𝑎𝑟𝑐𝑡𝑎𝑛(𝑒) − 𝑎𝑟𝑐𝑡𝑎𝑛(1)
1 𝑢 +1
π
●​ Answer: 𝑎𝑟𝑐𝑡𝑎𝑛(𝑒) − 4

→ 2^ ^ ^ → ^ ^ ^ 𝑑 → →
(f) If 𝑟1 = 𝑡 𝑖 − 𝑡𝑗 + (2𝑡 + 1) 𝑘, 𝑟2 = (2𝑡 − 3) 𝑖 + 𝑗 − 𝑡𝑘, find 𝑑𝑡
(𝑟1 · 𝑟2), where 𝑡 = 1.

●​ Key Idea: First calculate the dot product, then differentiate with respect to 𝑡, and finally substitute 𝑡 = 1.
→ → 2
●​ Steps: 𝑟1 · 𝑟2 = (𝑡 )(2𝑡 − 3) + (− 𝑡)(1) + (2𝑡 + 1)(− 𝑡)
3 2 2 3 2 𝑑 → → 𝑑 3 2 2
= 2𝑡 − 3𝑡 − 𝑡 − 2𝑡 − 𝑡 = 2𝑡 − 5𝑡 − 2𝑡 𝑑𝑡
(𝑟1 · 𝑟2) = 𝑑𝑡
(2𝑡 − 5𝑡 − 2𝑡) = 6𝑡 − 10𝑡 − 2
2
At 𝑡 = 1: 6(1) − 10(1) − 2 = 6 − 10 − 2 =− 6
●​ Answer: − 6
 → → →
(g) Write a necessary and sufficient condition for the three non-parallel, non zero vectors 𝑎, 𝑏, 𝑐 to be
coplanar.
●​ Key Idea: Three vectors are coplanar if their scalar triple product is zero.
→ → →
●​ Answer: The vectors 𝑎, 𝑏, 𝑐 are coplanar if and only if their scalar triple product is zero, i.e.,
→ → → → → →
𝑎 · (𝑏 × 𝑐) = 0. (Alternatively, [𝑎, 𝑏, 𝑐] = 0)
→
(h) Write the value of Div(ϕ𝑎).
●​ Key Idea: Use the product rule for divergence.
→ → → → → →
●​ Answer: 𝐷𝑖𝑣(ϕ𝑎) = ∇ · (ϕ𝑎) = (∇ϕ) · 𝑎 + ϕ(∇ · 𝑎) (or 𝑔𝑟𝑎𝑑(ϕ) · 𝑎 + ϕ𝑑𝑖𝑣(𝑎))
π π −10
(i) Simplify (𝑐𝑜𝑠 10
− 𝑖𝑠𝑖𝑛 10
) .

●​ Key Idea: Use De-Moivre's Theorem.

𝑛
●​ Steps: De-Moivre's Theorem states that (𝑐𝑜𝑠θ + 𝑖𝑠𝑖𝑛θ) = 𝑐𝑜𝑠(𝑛θ) + 𝑖𝑠𝑖𝑛(𝑛θ). We can rewrite the
π π −10
given expression as (𝑐𝑜𝑠(− 10
) + 𝑖𝑠𝑖𝑛(− 10
)) . Applying De-Moivre's Theorem:
π π
𝑐𝑜𝑠(− 10 · (− 10
)) + 𝑖𝑠𝑖𝑛(− 10 · (− 10
)) = 𝑐𝑜𝑠(π) + 𝑖𝑠𝑖𝑛(π) =− 1 + 𝑖(0)
●​ Answer: − 1
(j) Write the statement of De-Moivre's Theorem.
●​ Answer: De-Moivre's Theorem states that for any real number 𝑥 and integer 𝑛,
𝑛
(𝑐𝑜𝑠𝑥 + 𝑖𝑠𝑖𝑛𝑥) = 𝑐𝑜𝑠(𝑛𝑥) + 𝑖𝑠𝑖𝑛(𝑛𝑥). It also holds for rational numbers 𝑛 in a more general form.

𝑑𝑥
2. Integrate ∫ 2𝑠𝑖𝑛𝑥+3𝑐𝑜𝑠𝑥
.

●​ Method: We use the half-angle tangent substitution (Weierstrass substitution), where 𝑡 = 𝑡𝑎𝑛( 2𝑥 ).
2
2𝑡 1−𝑡 2𝑑𝑡
This implies 𝑠𝑖𝑛𝑥 = 2 , 𝑐𝑜𝑠𝑥 = 2 , and 𝑑𝑥 = 2 .
1+𝑡 1+𝑡 1+𝑡

2𝑑𝑡

Elaborate Steps: Substitute these into the integral: ∫

2

●​ 1+𝑡

( )+3( )
2
2𝑡 1−𝑡
2 2 2
1+𝑡 1+𝑡

2 2
4𝑡 3(1−𝑡 ) 4𝑡+3−3𝑡
​ Simplify the denominator: 2 + 2 = 2
1+𝑡 1+𝑡 1+𝑡

2𝑑𝑡
2
2𝑑𝑡
​ Now the integral becomes: ∫ 1+𝑡
3+4𝑡−3𝑡
2 = ∫ 2
2
3+4𝑡−3𝑡
1+𝑡

2𝑑𝑡 2 𝑑𝑡
​ Factor out − 3 from the denominator: ∫ 2 4
=− 3
∫ 2 4
−3(𝑡 − 𝑡−1) 3
𝑡 − 3 𝑡−1

2 2 2 2
​ Complete the square in the denominator: 𝑡 −
2 4
3
𝑡 − 1 =𝑡 −
2 4
3
𝑡+ ( ) −( )
3 3
− 1
2 2 2 2
(
= 𝑡 − 3) −
4
9
− 1= 𝑡 − ( 3 ) −
13
9
 2 𝑑𝑡
​ So the integral is: − 3
∫ 2
2 2
(𝑡− ) ( )
3
−
13
3

𝑑𝑥 1 𝑥−𝑎 2 13
​ This is of the form ∫ 2 2 = 2𝑎
𝑙𝑛|| 𝑥+𝑎 || + 𝐶. Here, 𝑥 = 𝑡 − 3
and 𝑎 = 3
.
𝑥 −𝑎

2 1 | 𝑡− 23 − 13
| 2 3 | 3𝑡−2− 13 | 1 | 3𝑡−2− 13 |
​ − · 𝑙𝑛|| 2 3 | + 𝐶 =−
| · 𝑙𝑛| | + 𝐶 =− 𝑙𝑛| |+ 𝐶
3
( )
2
13
3 | 𝑡− 3 +
13
3 | 3 2 13 | 3𝑡−2+ 13 | 13 | 3𝑡−2+ 13 |

𝑥
​ Substitute back 𝑡 = 𝑡𝑎𝑛( 2 ):

| 3𝑡𝑎𝑛( 2𝑥 )−2− |
Answer: − 1 13
●​ 𝑙𝑛| 𝑥 |+ 𝐶
13 | 3𝑡𝑎𝑛( 2 )−2+ 13 |

2
𝑥 𝑑𝑥
3. Integrate ∫ 2 .
(𝑥+1)(𝑥+2)

●​ Method: Use partial fraction decomposition. The denominator has a linear factor (𝑥 + 1) and a
2
repeated linear factor (𝑥 + 2) .

Elaborate Steps: Set up the partial fraction decomposition:

2
𝑥 𝐴 𝐵 𝐶
●​ 2 = 𝑥+1
+ 𝑥+2
+ 2
(𝑥+1)(𝑥+2) (𝑥+2)

2 2 2
​ Multiply both sides by (𝑥 + 1)(𝑥 + 2) : 𝑥 = 𝐴(𝑥 + 2) + 𝐵(𝑥 + 1)(𝑥 + 2) + 𝐶(𝑥 + 1)

​ To find the constants A, B, C:

​ Find A: Let 𝑥 =− 1: (− 1)2 = 𝐴(− 1 + 2)2 + 𝐵(− 1 + 1)(− 1 + 2) + 𝐶(− 1 + 1)

2
1 = 𝐴(1) + 0 + 0 𝐴 = 1

​ Find C: Let 𝑥 =− 2: (− 2)2 = 𝐴(− 2 + 2)2 + 𝐵(− 2 + 1)(− 2 + 2) + 𝐶(− 2 + 1)

4 = 0 + 0 + 𝐶(− 1) 𝐶 =− 4

​ Find B: Expand the equation or pick another convenient value for 𝑥, say 𝑥 = 0:
2 2
0 = 𝐴(0 + 2) + 𝐵(0 + 1)(0 + 2) + 𝐶(0 + 1) 0 = 4𝐴 + 2𝐵 + 𝐶 Substitute 𝐴 = 1 and
𝐶 =− 4: 0 = 4(1) + 2𝐵 + (− 4) 0 = 4 + 2𝐵 − 4 0 = 2𝐵 𝐵 = 0
2
𝑥 1 0 −4
​ Now substitute A, B, C back into the partial fraction form: 2 = 𝑥+1
+ 𝑥+2
+ 2
(𝑥+1)(𝑥+2) (𝑥+2)
1 4
= 𝑥+1
− 2
(𝑥+2)

​ Now integrate each term: ∫ ( 1

𝑥+1
−
(𝑥+2)
4
2 )𝑑𝑥 = ∫ 1
𝑥+1
𝑑𝑥 − 4∫(𝑥 + 2) 𝑑𝑥
−2

𝑛+1
𝑛 𝑢
​ The first integral is standard: 𝑙𝑛|𝑥 + 1|. For the second integral, use power rule: ∫𝑢 𝑑𝑢 = 𝑛+1
.
−2+1 −1
(𝑥+2) (𝑥+2) −1 4
− 4 −2+1
+ 𝐶 =− 4 −1
+ 𝐶 = 4(𝑥 + 2) +𝐶 = 𝑥+2
+ 𝐶

●​ Answer: 𝑙𝑛|𝑥 + 1| + 4
𝑥+2
+ 𝐶