IB Mathematics AA SL Indefinite integral of xn , sinx, cosx, and ex Study Notes - New Syllabus
IB Mathematics AA SL Indefinite integral of xn , sinx, cosx, and ex Study Notes
LEARNING OBJECTIVE
- Indefinite integral
Key Concepts:
- Indefinite integral
- Composite Function
- Integration by inspection
- IBDP Maths AA SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IBDP Maths AA SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Maths AA HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Maths AA HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Maths AA HL- IB Style Practice Questions with Answer-Topic Wise-Paper 3
Indefinite Integrals of Standard Functions
Indefinite Integrals of Standard Functions
Indefinite integration refers to finding the general antiderivative (primitive function) of a given function. These do not include limits and always include a constant of integration \( C \).
| Function \( f(x) \) | Antiderivative \( \int f(x) \, dx \) |
|---|---|
| \( x^n \), \( n \ne -1 \) | \( \dfrac{x^{n+1}}{n+1} + C \) |
| \( \dfrac{1}{x} \) | \( \ln |x| + C \) |
| \( e^x \) | \( e^x + C \) |
| \( \sin x \) | \( -\cos x + C \) |
| \( \cos x \) | \( \sin x + C \) |
Example
Find the integral of \( f(x) = 3x^2 + 4\cos x – \dfrac{2}{x} \)
▶️ Answer/Explanation
Solution:
Integrate term by term using standard formulas:
$ \int (3x^2 + 4\cos x – \dfrac{2}{x})\, dx = 3 \int x^2\, dx + 4 \int \cos x\, dx – 2 \int \dfrac{1}{x}\, dx$
$= 3 \cdot \dfrac{x^3}{3} + 4 \sin x – 2 \ln |x| + C $
Final Answer: \( x^3 + 4\sin x – 2\ln|x| + C \)
Integrating Composites with Linear Terms
Integrating Composites with Linear Terms
When integrating expressions of the form \( f(ax + b) \), use the rule:
\( \int f(ax + b)\, dx = \dfrac{1}{a}F(ax + b) + C \)
where \( F \) is the antiderivative of \( f \).
Example
Evaluate \( \displaystyle \int \cos(2x + 3)\, dx \)
▶️ Answer/Explanation
Solution:
\( \int \cos(2x + 3) \, dx = \dfrac{1}{2} \sin(2x + 3) + C \)
Example
Evaluate \( \int e^{3x + 2} \, dx \)
▶️ Answer/Explanation
Solution:
Let \( u = 3x + 2 \), so \( \frac{du}{dx} = 3 \Rightarrow dx = \frac{du}{3} \)
Substitute into the integral: \[ \int e^{3x + 2} \, dx = \int e^u \cdot \frac{1}{3} \, du = \frac{1}{3} \int e^u \, du = \frac{1}{3}e^u + C \]
Replace \( u \) with \( 3x + 2 \):
Final Answer: \( \frac{1}{3}e^{3x + 2} + C \)
Integration by Inspection or Substitution
Integration by Inspection or Substitution
Use this when the integrand resembles a chain rule derivative:
\( \int k g'(x) f(g(x)) \, dx \)
Steps for Substitution:
- Let \( u = g(x) \)
- Then \( du = g'(x)\, dx \)
- Rewrite the integral in terms of \( u \)
- Integrate and substitute back \( x \)
Example
Evaluate \( \int 2x(x^2 + 1)^4 \, dx \)
▶️ Answer/Explanation
Solution:
Recognize the inner function:
Let \( u = x^2 + 1 \Rightarrow \frac{du}{dx} = 2x \)
This matches the 2x outside:
$ \int 2x(x^2 + 1)^4 \, dx = \int (x^2 + 1)^4 \cdot 2x \, dx = \int u^4 \, du = \frac{u^5}{5} + C $
Example
Evaluate \( \int \dfrac{\sin x}{\cos x} \, dx \)
▶️ Answer/Explanation
Solution:
Let \( u = \cos x \Rightarrow du = -\sin x \, dx \)
Then: $ \int \frac{\sin x}{\cos x} \, dx = -\int \frac{1}{u} \, du = -\ln |u| + C = -\ln |\cos x| + C $
Example
Evaluate \( \int 4x\sin^2 x \, dx \)
▶️ Answer/Explanation
Use the identity \( \sin^2 x = \dfrac{1 – \cos(2x)}{2} \):
$ \int 4x\sin^2 x \, dx = \int 4x \cdot \frac{1 – \cos(2x)}{2} \, dx = \int 2x(1 – \cos(2x)) \, dx $
Split the integral:
$ \int 2x \, dx – \int 2x\cos(2x) \, dx $ First term: \( \int 2x \, dx = x^2 \)
Use integration by parts:
Let \( u = x \), \( dv = \cos(2x)\, dx \) ⇒ \( du = dx \), \( v = \dfrac{1}{2}\sin(2x) \)
$ \int 2x\cos(2x)\, dx = 2 \left( x \cdot \frac{1}{2}\sin(2x) – \int \frac{1}{2}\sin(2x)\, dx \right) $ $ = x\sin(2x) + \frac{1}{2}\cos(2x) $
