Calculus

Furthermore

Official Guidance, clarification and syllabus links:

\( \int \frac{1}{x} \; dx = \ln{|x|} + C\)

Example of composites:

\( f'(x)=\cos(2x+3) \; \Rightarrow \; f(x)= \frac12 \sin(2x+3) + C \)

Examples of inspection and substitution:

\( \int 2x(x^2 + 1)^4 \; dx, \quad \int 4x\sin{x^2} \; dx, \quad \int \frac{\sin x}{\cos x} \; dx \)

Formula Booklet:

Standard integrals

\( \int \frac{1}{x} \; dx = \ln{|x|} + C\) \( \int \sin x \; dx = - \cos x + C\) \( \int \cos x \; dx = \sin x + C\) \( \int e^x \; dx = e^x + C\)

Integration by inspection, also known as the reverse chain rule, is a technique used to simplify the process of integration for certain functions. When faced with an integrand of the form \( \int k g'(x)f(g(x)) \; dx \), one can recognise that the derivative of some function is present, allowing for a more straightforward integration.

To apply this method, we look for a function whose derivative matches a part of the integrand. If \( u = g(x) \), then \( du = g'(x) \; dx \). By substituting these values into the integral, the expression often simplifies, making it easier to evaluate.

Example:

Consider the integral:

$$ \int x \sin(x^2) \; dx $$

Here, we can let \( u = x^2 \). This gives \( \frac{du}{dx} = 2x \) or \( du = 2x \; dx \). Making the substitution, we get:

$$ \int \frac{1}{2} \sin(u) \; du $$

This integral is now straightforward to evaluate:

$$ -\frac{1}{2} \cos(u) + C $$

Re-substituting for \( u \), we obtain the final result:

$$ -\frac{1}{2} \cos(x^2) + C $$

This video on Integration Rules is from Revision Village and is aimed at students taking the IB Maths AA SL/HL course.

This video on Integration by Substitution is from Revision Village and is aimed at students taking the IB Maths AA SL/HL course