Question

Solve $\int \frac{1}{\sqrt[]{9-25x^2}}dx$

• A. $\frac{1}{5}{\sin }^{-1}\left(\frac{5x}{3}\right)+C$
• B. ${\sin }^{-1}\left(\frac{5x}{3}\right)+C$
• C. $\frac{1}{5}{\sin }^{-1}\left(\frac{3x}{5}\right)+C$
• D. ${\sin }^{-1}\left(\frac{3x}{5}\right)+C$

Answer 1

The correct option is A $\frac{1}{5}{\sin }^{-1}\left(\frac{5x}{3}\right)+C$

We have,

$I=\int \frac{dx}{\sqrt{9-25x^2}}$

$I=\int \frac{dx}{5\sqrt{\frac{9}{25}-x^2}}$

$I=\frac{1}{5}\int \frac{dx}{\sqrt{{\left(\frac{3}{5}\right)}^2-x^2}}$

We know that

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

Therefore,

$I=\frac{1}{5}{\sin }^{-1}\left(\frac{x}{\frac{3}{5}}\right)+C$

$I=\frac{1}{5}{\sin }^{-1}\left(\frac{5x}{3}\right)+C$

Hence, this is the correct answer.