Inverse Trigonometry Question (1 Viewer)

jxballistic · Mar 24, 2013

I am having trouble with part (ii). I know you have to use part (i) but I dont know how to properly explain

Trebla · Mar 24, 2013

jxballistic said:
View attachment 27888
I am having trouble with part (ii). I know you have to use part (i) but I dont know how to properly explain

lol Dr Du, this came straight out a past Maths Ext2 HSC paper.

Note that

$\displaystyle\int_0^{\frac{1}{2}}dx\leq \displaystyle\int_0^{\frac{1}{2}}\dfrac{dx}{\sqrt{1-x^n}}\leq \displaystyle\int_0^{\frac{1}{2}}\dfrac{dx}{\sqrt{1-x^2}}$

To prove this, note that the domain we are integrating over is 0 to 0.5 so we only need to consider 0 < x < 0.5 so first note that for n > 2

$0 \leq x^n \leq x^2 \\\\\Rightarrow - x^2 \leq - x^n \leq 0 \\\\\Rightarrow 1 - x^2 \leq 1 - x^n \leq 1\\\\\Rightarrow \sqrt{1 - x^2} \leq \sqrt{1 - x^n} \leq 1\\\\\Rightarrow 1 \leq \dfrac{1}{1-x^n} \leq \dfrac{1}{\sqrt{1-x^2}}$

then integrating both sides of the inequalities with the appropriate limits leads to the required result

jxballistic · Mar 24, 2013

Thanks, youre a great explainer. Very logical working out

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Inverse Trigonometry Question (1 Viewer)

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