Question

∫dx√2ax-x2=an sin-1[xa-1].
The value of n is
(a) 0
(b) −1
(c) 1
(d) none of these.
You may use dimensional analysis to solve the problem.

Answer 1

Solving the problem ∫dx√2ax-x2=an sin-1[xa-1] ,gives the value of n is 0.  

Explanation:

Given ∫dx √2ax-x2=an sin-1[xa-1].

To integrate the given equation,

                        $\text { Given } \int \frac{\mathrm{dx}}{\sqrt{\left(2 \mathrm{ax}-\mathrm{x}^{2}\right)}}$

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

                                  $=\int \frac{d x}{\sqrt{\left(a^{2}-\left(a^{2}-2 a x+x^{2}\right)\right.}}$

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

                               $\int \frac{\mathrm{dx}}{\sqrt{\left(2 \mathrm{ax}-\mathrm{x}^{2}\right)}}$   $=\sin ^{-1}\left[\frac{(\mathrm{x}-\mathrm{a})}{\mathrm{a}}\right]$    

Now comparing $a^n$ = 1 when n = 0.  

Substitute the values on both the sides of the equation and let x = 0

                                         0 $=\sin ^{-1}[0]$

After solving the values it gives,

LHS  value = 0 then RHS value = $sin^-^1$ (0) = 0.

(LHS = LEFT HAND SIDE) (RHS = RIGHT HAND SIDE)

The value of n is zero .

Answer 2

${\bold{\huge{\red{\underline{\green{ANSWER}}}}}}$

option A is correct