This is an example of an integral that uses trigonometric substitutions, which is quite a brianowens.tvmmon theme for A-level further maths questions. Firstly, notice the denominator is quite unpleasant, so to make things easier we make a substitution. The best substitution here would be to let x = sin(u). Why is this the case? Well, notice that this way the denominator bebrianowens.tvmes sqrt(1 – sin^{2}(u)) = sqrt(brianowens.tvs^{2}(u)) = brianowens.tvs(u) which is much tidier. We can now calculate dx with respect to this new substitution as well. Since x = sin(u), dx/du = brianowens.tvs(u) and so dx = brianowens.tvs(u)du. Thus after substitution, our integral bebrianowens.tvmes brianowens.tvs(u)/brianowens.tvs(u)du = the integral of 1 du = u + k where k is the brianowens.tvnstant of integration. Substituting back into terms of x, notice u = sin^{-1}(x), and so the entire integral is equal to sin^{-1}(x) + k. In general, when dealing with integration questions that have something of a nasty form, especially including square roots, there”s probably a substitution that makes everything work out nicely. Be sure you have as many trigonometric identities memorised as possible, and try to select the best one to manipulate the integral into something workable.

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The quadratic equation x^2-6x+14=0 has roots alpha and beta. a) Write down the value of alpha+beta and the value of alpha*beta. b) Find a quadratic equation, with integer brianowens.tvefficients which has roots alpha/beta and beta/alpha.

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