∫ √(1-49 x^2) d x is equal to

$\int \sqrt{1- (7x)^2} d x$

Let y = 7x  so dy = 7dx

⇒  $\frac{dy}{7} = dx$ 

So $I = \frac{1}{7} \int \sqrt{1-y^2} dy$

$=\frac{1}{7} \left[\frac{y}{2} \sqrt{1-y^2} + \frac{1}{2} \sin^{-1} y + C\right]$

$\Rightarrow \frac{1}{7} \times \frac{7x}{2} \sqrt{1 - 49x^2}+\frac{1}{7 \times 2} \sin ^{-1} 7x+C$

$=\frac{x}{2} \sqrt{1-49 x^2}+\frac{1}{14} \sin ^{-1} 7x+C$

Option: 4