Hard-difficulty flashcards covering advanced integration techniques, parametric differentiation, and trigonometric identities for Cambridge International A-Level Mathematics.
20 cards
Front
What is the derivative of the inverse tangent function, y = arctan(x)?
Back
dy/dx = 1 / (1 + x^2). This is derived by rewriting y = arctan(x) as x = tan(y) and using implicit differentiation with respect to x: 1 = sec^2(y) * dy/dx. Since sec^2(y) = 1 + tan^2(y) = 1 + x^2, we get dy/dx = 1/(1 + x^2).
Front
How do you integrate expressions of the form sin^n(x)?
Back
Use reduction formulas or trigonometric identities. For even powers (e.g., sin^2(x)), use the double-angle identity sin^2(x) = (1 - cos(2x))/2. For odd powers, separate one sin(x) factor and convert the remaining even power to cosines using sin^2 + cos^2 = 1.
Front
What is the method of Integration by Parts?
Back
Based on the product rule for differentiation: ∫ u (dv/dx) dx = uv - ∫ v (du/dx) dx. Choose u (to differentiate) and dv/dx (to integrate) strategically. A common mnemonic is LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) to prioritize u.
Front
How do you integrate a function of the form f'(x)/f(x)?
Back
The result is ln|f(x)| + c. This is because the derivative of ln(f(x)) is f'(x)/f(x) by the chain rule. For example, ∫ (2x / (x^2 + 1)) dx = ln|x^2 + 1| + c.
Front
What is the Weierstrass Substitution (t-substitution) used for?
Back
It is used to integrate rational functions of sine and cosine. It involves substituting t = tan(x/2). Consequently, sin(x) = 2t / (1 + t^2), cos(x) = (1 - t^2) / (1 + t^2), and dx = 2 / (1 + t^2) dt.
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