I was just doing revision for an upcoming exam and I came across a question I do not know how to answer. It seems pretty simple, however I am having a blank moment. If anyone knows how to solve that I would really appreciate the help.
The question:
It is given that y = $\sin^{-1}(ax)$, where a is a positive constant and $-\frac{1}{a}\leq x\leq \frac{1}{x}$.
- a) Find $\frac{dy}{dx}$ in terms of y and a.
- b) Hence find $\frac{dy}{dx}$ in terms of x and a.
- c) Hence, or otherwise, find $\int \frac{1}{\sqrt{1-9x^{2}}}dx$.
- d) Find the exact value of $\int_{0}^{1/6}\sqrt{1-9x^{2}} dx$
I have finished parts a to c (workings below) but cannot seem to solve part d (I do realize it's probably relatively simple, however I have just spent good 45 minutes trying to solve it or find the answer and I can't seem to solve it)
$ y=\sin^{-1}(ax)\\\sin(y)=ax\\\frac{dy}{dx}\cos(y)=a\\\Rightarrow\frac{dy}{dx}=\frac{a}{\cos(y)}=\frac{a}{\sqrt{1-\sin^{2}(y)}}=\frac{a}{\sqrt{1-a^{2}x^{2}}}\\\int\frac{1}{\sqrt{1-9x^{2}}}dx = \frac{1}{3}\int{\frac{3}{\sqrt{1-9x^{2}}}}dx=\frac{1}{3}\sin^{-1}(3x)+c$
Any help would be greatly appreciated.