Question

$\int \frac{dx}{\sqrt{2ax-x^2}}=a^n{\sin }^{-1}[\frac{x}{a}-1]$
The value of $n$ is:

• A. $0$
• B. $-1$
• C. $1$
• D. none of these

Answer 1

The correct option is A $0$

$LetI=\int \frac{1}{\sqrt{2ax-x^2}}dx=a^{n}si{n}^{-1}\left[\begin{array}{c}\frac{x}{2}-1\end{array}\right]+C$........(1)

= $\int \frac{1}{\sqrt{-(x^2-2ax)}}dx$

= $\int \frac{1}{\sqrt{-(x^2-2ax+a^2-a^2)}}dx$

= $\int \frac{1}{\sqrt{-\left(\right(x-a{)}^2-a^2)}}dx$

$\Rightarrow \int \frac{1}{\sqrt{a^2-(x-a{)}^2}}dx$

$\left[\begin{array}{c}\because \int \frac{1}{\sqrt{a^2-x^2}}dx=si{n}^{-1}\left[\begin{array}{c}\frac{x}{a}\end{array}\right]+C\end{array}\right]$

$\therefore$

$I=si{n}^{-1}\left[\begin{array}{c}\frac{x-a}{a}\end{array}\right]$

$I=a^{0}si{n}^{-1}\left[\begin{array}{c}\frac{x}{a}-1\end{array}\right]+C$........(2)

On comparing (1)&(2) we get

$a^n=a^0$

$\therefore \text{n=0 Hence correct opton is A}$