Question

Assertion :$\int \frac{dx}{\sqrt{1-(x^2{)}^2}}={\sin }^{-1}{x}^2+C$ Reason: $\int \frac{dx}{\sqrt{1-x^2}}={\sin }^{-1}x+C$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

Let ${\sin }^{-1}\left(x\right)=t$

$x=\sin t$

Hence $\cos t=\sqrt{1-x^2}$ ...(i)

$dx=\cos t.dt$

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

Now $t={\sin }^{-1}\left(x\right)$.

Hence

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

Or

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

Hence reason is correct.

Similarly

$y={\sin }^{-1}\left(x^2\right)$

$\sin \left(y\right)=x^2$

$\sqrt{\sin \left(y\right)}=x$

$\frac{dy}{dx}.\frac{\cos y}{2\sqrt{\sin y}}=1$

$\frac{dy}{dx}=\frac{2\sqrt{\sin y}}{\cos y}$

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

Or

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

Or

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

Hence assertion is incorrect.