Given that
y=secθ y = \sec \theta y=secθExpress yyy in terms of cosθ\cos \thetacosθ.
Hence, show that
dydθ=secθtanθ\frac{dy}{d\theta} = \sec \theta \tan \theta dθdy=secθtanθShow that for 0<θ<π20 < \theta < \frac{\pi}{2}0<θ<2π,
y2−1y=sinθ\frac{\sqrt{y^2-1}}{y} = \sin \theta yy2−1=sinθUse the substitution x=3secux = 3 \sec ux=3secu to show that for x>3x > 3x>3, the integral
∫1x2x2−9 dx\int \frac{1}{x^2 \sqrt{x^2 - 9}} \, dx ∫x2x2−91dxcan be written as
k∫cosu duk \int \cos u \, du k∫cosuduwhere kkk is a constant to be found.
Hence, show that
∫1x2x2−9 dx=x2−99x+C\int \frac{1}{x^2 \sqrt{x^2 - 9}} \, dx = \frac{\sqrt{x^2 - 9}}{9x} + C ∫x2x2−91dx=9xx2−9+CPractise Edexcel A Level Maths Integration with exam-style questions for A Level Maths. 437 questions covering 11.1 Integrating Standard Functions, 11.2 Integrating f(ax + b), 11.3 Using Trigonometric Identities, 11.4 Reverse Chain Rule, 11.5 Integration by Substitution, 11.6 Integration by Parts, 11.7 Partial Fractions, 11.8 Finding Areas, 11.9 The Trapezium Rule, 11.10 Solving Differential Equations, and 11.11 Modelling with Differential Equations, matched to the Edexcel A Level Maths (9MA0) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.
