# Solution-to-Quiz-1

```Solution to Quiz 1
SET A
1
1
3
1. ∫(8𝑥 4/5 − 3𝑥 5 + 2) √𝑥 2 𝑑𝑥 = ∫(8𝑥 4/5 − 3𝑥 5 + 2) 𝑥 2/3 𝑑𝑥
1
= 8 ∫ 𝑥 22/15 𝑑𝑥 − 3 ∫ 𝑥 −13/3 𝑑𝑥 + 2 ∫ 𝑥 2/3 𝑑𝑥
15
1
3
3
= 8⦁ 37 𝑥 37/15 − 3 (− 10) 𝑥 −10/3 + 2 (5) 𝑥 5/3 + 𝐶
=
120
37
1
6
𝑥 37/15 + 10𝑥 10/3 + 5 𝑥 5/3 + 𝐶
1
2. ∫ csc(5𝑥 + 3) cot(5𝑥 + 3)𝑑𝑥 = − csc(5𝑥 + 3) + 𝐶
5
3. ∫ sec 2 𝑥 (1 + tan 𝑥) 𝑑𝑥 = ∫ 𝑢𝑑𝑢
Let 𝑢 = 1 + tan 𝑥
𝑑𝑢 = sec 2 𝑥 𝑑𝑥
1
= 2 𝑢2 + 𝐶
1
= 2 (1 + tan 𝑥)2 + 𝐶
4. ∫ 𝑥 5 √2 − 𝑥 3 𝑑𝑥 = ∫ 𝑥 3 √2 − 𝑥 3 𝑥 2 𝑑𝑥
1
= − 3 ∫(2 − 𝑢)𝑢1/2 𝑑𝑢
2
1
= − ∫ 𝑢1/2 𝑑𝑢 + ∫ 𝑢3/2 𝑑𝑢
3
3
2 2
1 2
= − 3 (3) 𝑢3/2 + 3 (5) 𝑢5/2 + 𝐶
4
2
= − 9 (2 − 𝑥 3 )3/2 + 15 (2 − 𝑥 3 )5/2 + 𝐶
𝑢 = 2 − 𝑥3
𝑑𝑢 = −3𝑥 2 𝑑𝑥
1
− 3 𝑑𝑢 = 𝑥 2 𝑑𝑥
Let
𝑥3 = 2 − 𝑢
SET B
1. ∫ (3𝑥 4/3 −
1
2𝑥 4
1
+ 5) √𝑥 3 𝑑𝑥 = ∫(3𝑥 4/3 − 2𝑥 4 + 5) 𝑥 3/2 𝑑𝑥
1
= 3 ∫ 𝑥 17/6 𝑑𝑥 − 2 ∫ 𝑥 −5/2 𝑑𝑥 + 5 ∫ 𝑥 3/2 𝑑𝑥
6
1
2
2
= 3⦁ 23 𝑥 23/6 − 2 (− 3) 𝑥 −3/2 + 5 (5) 𝑥 5/2 + 𝐶
18
1
= 23 𝑥 23/6 + 3𝑥 3/2 + 2𝑥 5/2 + 𝐶
1
2. ∫ sec 2 (2 − 5𝑥)𝑑𝑥 = − 5 tan(2 − 5𝑥) + 𝐶
3. ∫(1 + cot 𝑥) csc 2 𝑥 𝑑𝑥 = − ∫ 𝑢𝑑𝑢
Let 𝑢 = 1 + cot 𝑥
𝑑𝑢 = −csc 2 𝑥 𝑑𝑥
1
= − 2 𝑢2 + 𝐶
1
= − 2 (1 + cot 𝑥)2 + 𝐶
4. ∫ 𝑥 3 √2 − 𝑥 2 𝑑𝑥 = ∫ 𝑥 2 √2 − 𝑥 2 𝑥𝑑𝑥
1
= − 2 ∫(2 − 𝑢)𝑢1/2 𝑑𝑢
1
= − ∫ 𝑢1/2 𝑑𝑢 + 2 ∫ 𝑢3/2 𝑑𝑢
2
1 2
= − 3 𝑢3/2 + 2 (5) 𝑢5/2 + 𝐶
2
1
= − 3 (2 − 𝑥 2 )3/2 + 5 (2 − 𝑥 2 )5/2 + 𝐶
𝑢 = 2 − 𝑥2
𝑑𝑢 = −2𝑥𝑑𝑥
1
− 2 𝑑𝑢 = 𝑥𝑑𝑥
Let
𝑥2 = 2 − 𝑢
```