Match List-I with List-II (Given that $c$ is an arbitrary constant)

List-I	List-II
(A) $\int \frac{dx}{\sqrt{a^2 - x^2}} =$	(I) $\log_e \vert x + \sqrt{x^2 - a^2}\vert + c$
(B) $\int \sqrt{a^2 - x^2} dx =$	(II) $\sin^{-1} \frac{x}{a} + c$
(C) $\int \sqrt{x^2 - a^2} dx =$	(III) $\frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \frac{x}{a} + c$
(D) $\int \frac{dx}{\sqrt{x^2 - a^2}} =$	(IV) $\frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \log_e \vert x + \sqrt{x^2 - a^2}\vert + c$

Choose the correct answer from the options given below:

These are standard integral formulas. Let's match each one by recalling the standard results.

$\int \dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\dfrac{x}{a} + c$

This is derived using the substitution $x = a\sin\theta$. The expression $(a^2 - x^2)$ under the root pairs with $\sin^{-1}$.

$(A) \rightarrow (II)$

$\int \sqrt{a^2 - x^2}\, dx = \dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + c$

Since the expression under the root is $(a^2 - x^2)$, the result contains $\sin^{-1}$ (not $\log_e$). Notice the $+$ sign between the two terms.

$(B) \rightarrow (III)$

$\int \sqrt{x^2 - a^2}\, dx = \dfrac{x}{2}\sqrt{x^2 - a^2} - \dfrac{a^2}{2}\log_e|x + \sqrt{x^2 - a^2}| + c$

Since the expression under the root is $(x^2 - a^2)$, the result contains $\log_e$ (not $\sin^{-1}$). Notice the $-$ sign between the two terms.

$(C) \rightarrow (IV)$

$\int \dfrac{dx}{\sqrt{x^2 - a^2}} = \log_e|x + \sqrt{x^2 - a^2}| + c$

This is the simpler form where $(x^2 - a^2)$ is under the root in the denominator, giving a direct $\log_e$ result.

$(D) \rightarrow (I)$

Memory tip: $\sin^{-1}$ always pairs with $(a^2 - x^2)$ type integrals, and $\log_e$ always pairs with $(x^2 - a^2)$ type integrals.

$(A) \rightarrow (II),\; (B) \rightarrow (III),\; (C) \rightarrow (IV),\; (D) \rightarrow (I)$