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+C
