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+C864 exam-style questions on Edexcel A Level Maths Integration, 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. Each one has a worked solution and a mark scheme showing where the marks go.