# (iüi) sin 3x (iv) Jax + b (iv) sin(2x+3) (vii) sec 3x tan 3x (i) co 2x 9 cosee(1 -31)

**URL:** <https://sorumatik.co/t/iui-sin-3x-iv-jax-b-iv-sin-2x-3-vii-sec-3x-tan-3x-i-co-2x-9-cosee-1-31/690983>\
**Category:** Maths Question\
**Created:** [Eylül 6, 2026, 5:36ös UTC](https://sorumatik.co/t/iui-sin-3x-iv-jax-b-iv-sin-2x-3-vii-sec-3x-tan-3x-i-co-2x-9-cosee-1-31/690983 "2026-09-06T17:36:29Z")\
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**Post date:** [Eylül 6, 2026, 5:36ös UTC](https://sorumatik.co/t/iui-sin-3x-iv-jax-b-iv-sin-2x-3-vii-sec-3x-tan-3x-i-co-2x-9-cosee-1-31/690983/1 "2026-09-06T17:36:29Z")

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1. (0((-2x)  
(),(v) (1-4xl2  
(vi) J+x  
(i3) J1+x
2. () e  
(ii) e
3. () sinmx  
(iüi) sin 3x  
(v)

\cos(2x-3)

(vii) cos2x  
() (4+5\*}  
(iv) Jax + b  
() sin 2x+)  
(vii)  
3-5x  
(i) e\*  
() \sin  
(iv)

\sin(2x+3)

(vi)

\cos

(vii) sec 3x tan 3x  
(i) co 2x 9 cosee(1 -31)

 ![(0((-2x) (),(v) (1-4xl2 (vi) J+x (i3) J1+x 2.](https://sorumatik.co/uploads/default/original/3X/f/1/f122d37b59e7bdbc90275b92ba433f1969f0b0cb.jpeg)

---

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**Author:** ![Egitim\_Danismani](https://sorumatik.co/user_avatar/sorumatik.co/egitim_danismani/32/86791_2.png) [@Egitim\_Danismani](https://sorumatik.co/u/Egitim_Danismani)\
**Post date:** [Eylül 6, 2026, 5:37ös UTC](https://sorumatik.co/t/iui-sin-3x-iv-jax-b-iv-sin-2x-3-vii-sec-3x-tan-3x-i-co-2x-9-cosee-1-31/690983/2 "2026-09-06T17:37:20Z")

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## Integration of Algebraic, Exponential, and Trigonometric Functions

💡 **FORMULAS USED:**

1. **Power Rule of Linear Function:**

2. **Logarithmic Rule:**

3. **Exponential Rule:**

4. **Trigonometric Rules:**

* * *

🧠 **SOLUTION STEPS:**

### Section 1: Algebraic Integrals

**Step 1 — Solve 1.(i) \int (1-2x)^4 \, dx**

- Apply the linear power rule where a = -2 and n = 4:
\int (1-2x)^4 \, dx = \frac{(1-2x)^5}{(-2) \times 5} + C = -\frac{(1-2x)^5}{10} + C

**Step 2 — Solve 1.(ii) \int (4+5x)^3 \, dx**

- Apply the linear power rule where a = 5 and n = 3:
\int (4+5x)^3 \, dx = \frac{(4+5x)^4}{5 \times 4} + C = \frac{(4+5x)^4}{20} + C

**Step 3 — Solve 1.(iii) \int \frac{1}{(1+x)^4} \, dx**

- Rewrite as \int (1+x)^{-4} \, dx where a = 1 and n = -4:
\int (1+x)^{-4} \, dx = \frac{(1+x)^{-3}}{-3} + C = -\frac{1}{3(1+x)^3} + C

**Step 4 — Solve 1.(iv) \int \sqrt{ax+b} \, dx**

- Rewrite as \int (ax+b)^{1/2} \, dx:
\int (ax+b)^{1/2} \, dx = \frac{(ax+b)^{3/2}}{a \times \frac{3}{2}} + C = \frac{2(ax+b)^{3/2}}{3a} + C

**Step 5 — Solve 1.(v) \int \frac{1}{2x+3} \, dx**

- Apply the logarithmic rule with a = 2:
\int \frac{1}{2x+3} \, dx = \frac{1}{2} \ln|2x+3| + C

**Step 6 — Solve 1.(vi) \int (1-4x)^{-12} \, dx**

- Apply the power rule where a = -4 and n = -12:
\int (1-4x)^{-12} \, dx = \frac{(1-4x)^{-11}}{(-4) \times (-11)} + C = \frac{1}{44(1-4x)^{11}} + C

**Step 7 — Solve 1.(vii) \int \frac{1}{\sqrt{1+x}} \, dx**

- Rewrite as \int (1+x)^{-1/2} \, dx:
\int (1+x)^{-1/2} \, dx = \frac{(1+x)^{1/2}}{\frac{1}{2}} + C = 2\sqrt{1+x} + C

**Step 8 — Solve 1.(viii) \int \frac{1}{3-5x} \, dx**

- Apply the logarithmic rule where a = -5:
\int \frac{1}{3-5x} \, dx = -\frac{1}{5} \ln|3-5x| + C

**Step 9 — Solve 1.(ix) \int \sqrt{1+x} \, dx**

- Rewrite as \int (1+x)^{1/2} \, dx:
\int (1+x)^{1/2} \, dx = \frac{(1+x)^{3/2}}{\frac{3}{2}} + C = \frac{2}{3}(1+x)^{3/2} + C

**Step 10 — Solve 1.(x) \int \frac{1}{(1-3x)^2} \, dx**

- Rewrite as \int (1-3x)^{-2} \, dx where a = -3 and n = -2:
\int (1-3x)^{-2} \, dx = \frac{(1-3x)^{-1}}{(-3) \times (-1)} + C = \frac{1}{3(1-3x)} + C

* * *

### Section 2: Exponential Integrals

**Step 11 — Solve 2.(i) \int e^{mx} \, dx**

- Apply the exponential rule:
\int e^{mx} \, dx = \frac{1}{m}e^{mx} + C

**Step 12 — Solve 2.(ii) \int e^{-x} \, dx**

- Apply the exponential rule where a = -1:
\int e^{-x} \, dx = -e^{-x} + C

**Step 13 — Solve 2.(iii) \int e^{3x} \, dx**

- Apply the exponential rule where a = 3:
\int e^{3x} \, dx = \frac{1}{3}e^{3x} + C

* * *

### Section 3: Trigonometric Integrals

**Step 14 — Solve 3.(i) \int \sin(mx) \, dx**

- Apply \int \sin(ax) \, dx = -\frac{1}{a}\cos(ax):
\int \sin(mx) \, dx = -\frac{1}{m}\cos(mx) + C

**Step 15 — Solve 3.(ii) \int \sin\left(\frac{x}{2}\right) \, dx**

- Here a = \frac{1}{2}:
\int \sin\left(\frac{x}{2}\right) \, dx = -\frac{1}{1/2}\cos\left(\frac{x}{2}\right) + C = -2\cos\left(\frac{x}{2}\right) + C

**Step 16 — Solve 3.(iii) \int \sin(3x) \, dx**

- Here a = 3:
\int \sin(3x) \, dx = -\frac{1}{3}\cos(3x) + C

**Step 17 — Solve 3.(iv) \int \sin(2x+3) \, dx**

- Here a = 2:
\int \sin(2x+3) \, dx = -\frac{1}{2}\cos(2x+3) + C

**Step 18 — Solve 3.(v) \int \cos(2x-3) \, dx**

- Apply \int \cos(ax+b) \, dx = \frac{1}{a}\sin(ax+b):
\int \cos(2x-3) \, dx = \frac{1}{2}\sin(2x-3) + C

**Step 19 — Solve 3.(vi) \int \cos\left(\frac{x}{2}\right) \, dx**

- Here a = \frac{1}{2}:
\int \cos\left(\frac{x}{2}\right) \, dx = \frac{1}{1/2}\sin\left(\frac{x}{2}\right) + C = 2\sin\left(\frac{x}{2}\right) + C

**Step 20 — Solve 3.(vii) \int \cos(2x) \, dx**

- Here a = 2:
\int \cos(2x) \, dx = \frac{1}{2}\sin(2x) + C

**Step 21 — Solve 3.(viii) \int \sec(3x)\tan(3x) \, dx**

- Apply standard form where a = 3:
\int \sec(3x)\tan(3x) \, dx = \frac{1}{3}\sec(3x) + C

**Step 22 — Solve 3.(ix) \int \cos\left(\frac{\pi}{6}+2x\right) \, dx**

- Here a = 2:
\int \cos\left(\frac{\pi}{6}+2x\right) \, dx = \frac{1}{2}\sin\left(\frac{\pi}{6}+2x\right) + C

**Step 23 — Solve 3.(x) \int \csc^2(1-3x) \, dx**

- Here a = -3:
\int \csc^2(1-3x) \, dx = -\frac{1}{-3}\cot(1-3x) + C = \frac{1}{3}\cot(1-3x) + C

**Step 24 — Solve 3.(xi) \int \sin\left(2x+\frac{\pi}{4}\right) \, dx**

- Here a = 2:
\int \sin\left(2x+\frac{\pi}{4}\right) \, dx = -\frac{1}{2}\cos\left(2x+\frac{\pi}{4}\right) + C

━━━━━━━━━━━━━━━━━━━━━━━━━━━  
✅ **FINAL ANSWERS SUMMARY:**

- **1.(i):** -\frac{1}{10}(1-2x)^5 + C
- **1.(ii):** \frac{1}{20}(4+5x)^4 + C
- **1.(iii):** -\frac{1}{3(1+x)^3} + C
- **1.(iv):** \frac{2}{3a}(ax+b)^{3/2} + C
- **1.(v):** \frac{1}{2}\ln|2x+3| + C
- **1.(vi):** \frac{1}{44(1-4x)^{11}} + C
- **1.(vii):** 2\sqrt{1+x} + C
- **1.(viii):** -\frac{1}{5}\ln|3-5x| + C
- **1.(ix):** \frac{2}{3}(1+x)^{3/2} + C
- **1.(x):** \frac{1}{3(1-3x)} + C
- **2.(i):** \frac{1}{m}e^{mx} + C
- **2.(ii):** -e^{-x} + C
- **2.(iii):** \frac{1}{3}e^{3x} + C
- **3.(i):** -\frac{1}{m}\cos(mx) + C
- **3.(ii):** -2\cos\left(\frac{x}{2}\right) + C
- **3.(iii):** -\frac{1}{3}\cos(3x) + C
- **3.(iv):** -\frac{1}{2}\cos(2x+3) + C
- **3.(v):** \frac{1}{2}\sin(2x-3) + C
- **3.(vi):** 2\sin\left(\frac{x}{2}\right) + C
- **3.(vii):** \frac{1}{2}\sin(2x) + C
- **3.(viii):** \frac{1}{3}\sec(3x) + C
- **3.(ix):** \frac{1}{2}\sin\left(2x+\frac{\pi}{6}\right) + C
- **3.(x):** \frac{1}{3}\cot(1-3x) + C
- **3.(xi):** -\frac{1}{2}\cos\left(2x+\frac{\pi}{4}\right) + C  
━━━━━━━━━━━━━━━━━━━━━━━━━━━

🎯 **KEY CONCEPTS:**

**1. Linear Substitution in Integration**

- **Definition:** When an integrand contains a linear term (ax+b) instead of x, the integral of f(ax+b) is \frac{1}{a}F(ax+b) + C, where F'(x) = f(x).
- **In this problem:** We divided each standard anti-derivative by the coefficient of x (the value of a).

⚠ **COMMON MISTAKES:**

**❌ Forgetting to divide by the coefficient of x**

- **Wrong:** \int \sin(3x) \, dx = -\cos(3x) + C
- **Right:** \int \sin(3x) \, dx = -\frac{1}{3}\cos(3x) + C
- **Why it’s wrong:** Differentiating -\cos(3x) by the chain rule gives 3\sin(3x), so the factor of \frac{1}{3} is required to cancel the 3.

**❌ Sign errors with negative x coefficients**

- **Wrong:** \int \frac{1}{3-5x} \, dx = \frac{1}{5} \ln|3-5x| + C
- **Right:** \int \frac{1}{3-5x} \, dx = -\frac{1}{5} \ln|3-5x| + C
- **Why it’s wrong:** The coefficient of x is -5, so the factor in front must be -\frac{1}{5}.

* * *

Would you like me to walk you through a step-by-step u-substitution proof for any specific problem from this set?