Question

I:$\int \frac{dx}{\sqrt{9-x^2}}={\sin }^{-1}\left(\frac{x}{3}\right)+c$
II:$\int \frac{\cos x}{\sqrt{16-{\sin }^{2}x}}dx$-- ${\sin }^{-1}\left(\frac{\sin x}{4}\right)+c$

• A. Only I
• B. Only II
• C. Both I and II
• D. neither I nor II are true

Answer 1

The correct option is C Both I and II

I:$\int \frac{dx}{\sqrt{9-x^2}}={\sin }^{-1}\left(\frac{x}{3}\right)+c$
Consider, $\int \frac{dx}{\sqrt{9-x^2}}$
$=\int \frac{dx}{\sqrt{3^2-x^2}}$
$={\sin }^{-1}\left(\frac{x}{3}\right)+c$
Statement 1 is true.
II:$\int \frac{\cos x}{\sqrt{16-{\sin }^{2}x}}dx$= ${\sin }^{-1}\left(\frac{\sin x}{4}\right)+c$
Consider, $\int \frac{\cos x}{\sqrt{16-{\sin }^{2}x}}dx$
Put $\sin x=t$
$\cos xdx=dt$
$I=\int \frac{dt}{\sqrt{16-t^2}}$
$={\sin }^{-1}\left(\frac{t}{4}\right)+c$
$={\sin }^{-1}\left(\frac{\sin x}{4}\right)+c$
Statement 2 is true.