∫sin^-1(2x) dx is equal to

The correct answer is Option (2) → \(\displaystyle x\sin^{-1}(2x)+\frac{1}{2}\sqrt{1-4x^2}+C\)

Let \(I = \int\sin^{-1}(2x)\cdot 1\,dx\). Using integration by parts with \(\sin^{-1}(2x)\) as the first function and 1 as the second, and \(\frac{d}{dx}\sin^{-1}(2x) = \frac{2}{\sqrt{1-4x^2}}\):

    \( \displaystyle I = x\sin^{-1}(2x)-\int\frac{2x}{\sqrt{1-4x^2}}\,dx \)

With \(t = 1-4x^2\), \(dt = -8x\,dx\), so \(2x\,dx = -\frac{dt}{4}\):

    \( \displaystyle \int\frac{2x}{\sqrt{1-4x^2}}\,dx = -\frac{1}{4}\int\frac{dt}{\sqrt{t}} = -\frac{1}{2}\sqrt{1-4x^2} \)

So

    \( \displaystyle I = x\sin^{-1}(2x)+\frac{1}{2}\sqrt{1-4x^2}+C \)

Why the other options are incorrect

• Option (1) \(\displaystyle x\sin^{-1}(2x)-\frac{1}{2}\sqrt{1-4x^2}+C\) — Incorrect. The sign of the root term is wrong; its derivative is \(\sin^{-1}(2x)+\frac{4x}{\sqrt{1-4x^2}}\).

• Option (3) \(x\sin^{-1}(2x)+2\sqrt{1-4x^2}+C\) — Incorrect. The root term should have the factor \(\frac{1}{2}\), not 2; its derivative is \(\sin^{-1}(2x)-\frac{6x}{\sqrt{1-4x^2}}\).

• Option (4) \(\displaystyle x\sin^{-1}(2x)+\frac{1}{2\sqrt{1-4x^2}}+C\) — Incorrect. \(\int\frac{dt}{\sqrt{t}}\) is \(2\sqrt{t}\), so the root must stay in the numerator.