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Re: HSC 2018 MX2 Integration Marathon 1989 last question?

Re: HSC 2018 MX2 Integration Marathon The next task is to find the indefinite integral. Of course the answer is not sin x-cos x+c. \int\sqrt{1+\sin(2x)}dx Hint: You may consider the floor function.

Re: HSC 2018 MX2 Integration Marathon This one may look simple at the first glance but actually trickier than you may have thought. I'm sure a lot of people will come up with an answer 2.:devil: \int_0^{\pi}\sqrt{1+\sin(2x)}dx

Re: HSC 2018 MX2 Integration Marathon This is a skeleton solution. By substituting u=(x-2)/sqrt(2) and considering f(x)+f(-x), the integral can be re-written as \frac{\sqrt{2}}{8}\int_0^{\sqrt{2}-1}\frac{\sqrt{1+u^4}}{1-u^4}du A tangent substitution will turn it into a format that...

Re: HSC 2018 MX2 Integration Marathon This is another beast.:cool2: \int_{4-\sqrt{2}}^{\sqrt{2}}\frac{\sqrt{(x^2-6x+10)(x^2-2x+2)}}{(x^2-4x+2)(x^2-4x+6)(4+2^x)}dx

Re: HSC 2018 MX2 Integration Marathon when f(x)+f(a-x) is easier to integrate than f(x)

Re: HSC 2018 MX2 Integration Marathon This one is absolutely a beast. \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\frac{(\sec x)\sqrt{3+\cos2x}}{1+2019^x}dx

Re: HSC 2018 MX2 Integration Marathon You are almost there. If you put x=tan theta, does it look familiar? You've solved that in the previous one.:devil:

Re: HSC 2018 MX2 Integration Marathon This one should be considerably easier than the previous one.:tongue: \int_{-1}^{1}\frac{x^{2018}\sqrt{1-\cos^2(\frac{\pi}{2}x^{2019})} \log_2(\sec(\frac{\pi}{4}x^{2019}))}{(3+\cos(\pi x^{2019}))(1+2018^x)}dx The answer is pretty small. (1/32304)

Re: HSC 2018 MX2 Integration Marathon Continue to have fun with trig.:jump: Harder version: f(x)=\frac{\sin(\frac{\pi}{2}\sqrt{x})}{\sqrt{2}+ \sqrt{2}\cos(\frac{\pi}{2}\sqrt{x})}+\frac{x\sec (\frac{\pi}{4})}{...

Re: HSC 2018 MX2 Integration Marathon Construct from simpler integrals using trig identities and properties of definite integrals.

Re: HSC 2018 MX2 Integration Marathon Just consider the integral from 0 to a first, then take the limit as a->pi-.

Re: HSC 2018 MX2 Integration Marathon Show that \int_{0}^{\frac{\pi}{2}} \frac{sin(x+\frac{\pi}{4})}{(2^\pi+16^x)(sin^3x+cos^3x)}dx=\frac{\sqrt{6}\pi}{9(2^\pi)}

Consider two antiderivatives below. f(x)=-1/x+2 for x>0 f(x)=-1/x for x<0 f'(x)=1/x^2 for all non-zero real values of x f(1)=1, f(-1)=1 f(x)=-1/x+2 for x>0 f(x)=-1/x+2 for x<0 f'(x)=1/x^2 for all non-zero real values of x f(1)=1, f(-1)=3 Both satisfy the requirement. Therefore...

Technically, the anti-derivative of 1/x^2 should be -1/x+C1 for x>0 -1/x+C2 for x<0 where C1 and C2 are (possibly different) constants. It definitely matters which side you are looking at. Suppose f'(x)=1/x^2 and f(1)=1. You cannot determine f(-1).

Technically, in the world of real functions, the anti-derivative of 1/x should be ln(x)+C1 for x>0 ln(-x)+C2 for x<0 where C1 and C2 are (possibly different) constants. When the question is only looking at either the positive or negative side, then it can be written as ln|x|+C, where C is a...

Re: HSC 2018 MX2 Integration Marathon Show that \int_{0}^{\frac{\pi}{4}}\frac{2\sin 8x \cos 4x \cos 2x}{(1+\cos 8x)(1+\cos 4x)(1+\cos 2x))\sqrt{2+\cos 2x}}dx =\ln(6+4\sqrt{2}-3\sqrt{3}-2\sqrt{6})

Re: HSC 2018 MX2 Integration Marathon This one is fairly straight-forward. \int (sec x-tan x)^{\frac{2018}{2017}}d(tan x)

Re: HSC 2018 MX2 Integration Marathon For x\geqslant 0, define f(x)=\int_{0}^{\pi}t\,sin^{x}t\,dt. (a)Show that f(x) is decreasing. Write down the maximum value of f(x). (b)Find \int \ln\,f(x) - \ln\,f(x+2)\,dx.

Re: HSC 2017 MX2 Integration Marathon Try this one. \cot 3x\sec^2(ln \sqrt{sin 3x})