# Evaluating integral similar to $\sin^{-1}(3x)$

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.

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For the last part, let $x = \dfrac{\sin(\theta)}{3}$. As $x$ goes from $0$ to $1/6$, we have that $\theta$ goes from $0$ to $\pi/6$. Also, $dx = \dfrac{\cos(\theta)}{3} d \theta$. Hence, $$I = \int_0^{1/6} \sqrt{1-9x^2} dx = \int_0^{\pi/6} \sqrt{1 - \sin^2(\theta)} \dfrac{\cos(\theta)}{3} d \theta = \int_0^{\pi/6} \dfrac{\cos^2(\theta)}{3} d \theta$$ Hence, $$I = \dfrac16 \int_0^{\pi/6} 2 \cos^2(\theta) d \theta = \dfrac16 \int_0^{\pi/6} (1 + \cos(2 \theta)) d \theta = \dfrac16 \left( \theta + \dfrac{\sin(2 \theta)}{2} \right)_{0}^{\pi/6}$$ Hence, $$I = \dfrac16 \left(\dfrac{\pi}{6} + \dfrac{\sin(\pi/3)}{2}\right) = \dfrac16 \left(\dfrac{\pi}{6} + \dfrac{\sqrt{3}}{4}\right) = \dfrac1{72} \left(2 \pi + 3 \sqrt{3} \right)$$

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Thanks a lot! :) This makes perfect sense, I can't believe I haven't spotted this way in all the time I spent trying to solve this problem. Lack of sleep is getting to me. Thanks again, you're a life savior! :) –  Bart Platak Jun 6 '12 at 2:00

use this formula $$\int \sqrt{a^2-x^2}dx=\frac{1}{2}x\sqrt{a^2-x^2}+\frac{a^2}{2}\arcsin \frac{x}{a}+C$$ $$\int_{0}^{1/6}\sqrt{1-9x^{2}} dx$$ $$\left(\frac{3x}{6}\sqrt{1-9x^{2}}+\frac{1}{6}\arcsin\frac{3x}{1} \right)_{0}^{1/6}$$ $$\left(\frac{x}{2}\sqrt{1-9x^{2}}+\frac{1}{6}\arcsin\frac{3x}{1} \right)_{0}^{1/6}$$ $$(\frac{1}{12}\sqrt{1-\frac{9}{36}}+\frac{1}{6}\arcsin\frac{1}{2})-0$$ $$\frac{\sqrt{27}}{72}+\frac{\pi}{36}$$ $$\dfrac1{72} \left(2 \pi + 3 \sqrt{3} \right)$$

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