1. ## Inverse trig

Differentiate $sin^{ - 1}[\frac{(x-5)}{5}]$ and hence evaulate $\int_5^{10} \frac{dx}{\sqrt (10x-x^2)}$

I'm not sure if the number in front of the x in the square root is a 10 or a 30.

2. Hello, nerdzor!

Have you done anything yet?

Differentiate $\sin^{\text{-}1}\!\left(\frac{x-5}{5}\right)$ . and hence evaulate: . $\int_5^{10} \frac{dx}{\sqrt{10x-x^2}}$
We have: . $y \:=\:\sin^{-1}\left(\frac{x-5}{5}\right)$

Differentiate: . $y' \;=\;\frac{\frac{1}{5}}{\sqrt{1-\left(\frac{x-5}{5}\right)^2}} \;=\;\frac{\frac{1}{5}}{\sqrt{1 - \frac{(x-5)^2}{25}}} \;=\;\frac{\frac{1}{5}}{\sqrt{\frac{25-(x-5)^2}{25}}}$

. . $y' \;=\;\frac{\frac{1}{5}}{\frac{\sqrt{25-(x-5)^2}}{5}} \;=\;\frac{1}{\sqrt{25 - x^2 + 10x - 25}} \;=\;\frac{1}{\sqrt{10x - x^2}}$

Get it?

The derivative of $\sin^{-1}\left(\frac{x-5}{5}\right)$ is: . $\frac{1}{\sqrt{10x-x^2}}$

So the integral of $\frac{1}{\sqrt{10x - x^2}}$ is: . $\sin^{\text{-}1}\left(\frac{x-5}{5}\right) + C$

Hence: . $\int^{10}_5\frac{dx}{\sqrt{10x - x^2}} \;=\;\sin^{\text{-}1}\left(\frac{x-5}{5}\right)\,\bigg]^{10}_5 \;=\;\sin^{\text{-}1}\left(\frac{10-5}{5}\right) - \sin^{\text{-}1}\left(\frac{5-5}{5}\right)$

. . . . . . $= \;\; \sin^{\text{-}1}(1) - \sin^{\text{-}1}(0) \;\;=\;\;\frac{\pi}{2} - 0 \;\;=\;\;\frac{\pi}{2}$

3. Originally Posted by nerdzor
Differentiate $sin^{ - 1}[\frac{(x-5)}{5}]$ and hence evaulate $\int_5^{10} \frac{dx}{\sqrt (10x-x^2)}$

I'm not sure if the number in front of the x in the square root is a 10 or a 30.
It's a $10$. I know because here's the answer to the first part. You then use this to find the integral.

$y = \sin^{-1}\Big(\frac{x-5}{5}\Big)$

$\Rightarrow \sin y = \frac{x-5}{5}$

$\Rightarrow \cos y \frac{dy}{dx} = \frac15$

$\Rightarrow \frac{dy}{dx} = \frac{1}{5\cos y}$

$= \frac{1}{5\sqrt{1 - \sin^2y}}$

$=\frac {1}{\sqrt{25- 25\sin^2y}}$

$=\frac{1}{\sqrt{25 - (x-5)^2}}$

$=\frac{1}{\sqrt{10x - x^2}}$

Can you complete the second part now?