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MHT CET · Maths · Indefinite Integration

If \(x \in[-1,1]\), then the value of \(\int \mathrm{e}^{\sin ^{-1} x}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{1-x^2}}\right) \mathrm{d} x\) is

  1. A \(\mathrm{e}^{\sin ^{-1} x}+\mathrm{c}\), where c is constant of integration.
  2. B \(\mathrm{e}^{\sin ^{-1} x} \cdot \sin x+\mathrm{c}\), where c is constant of integration.
  3. C \(\mathrm{e}^{\sin ^{-1} x} \cdot \cos x+\mathrm{c}\), where c is constant of integration.
  4. D \(\mathrm{e}^{\sin ^{-1} x} \cdot x+\mathrm{c}\), where c is constant of integration.
Verified Solution

Answer & Solution

Correct Answer

(D) \(\mathrm{e}^{\sin ^{-1} x} \cdot x+\mathrm{c}\), where c is constant of integration.

Step-by-step Solution

Detailed explanation

\(\begin{aligned}
& \text { Put } \sin ^{-1} x=\mathrm{t} \Rightarrow \frac{1}{\sqrt{1-x^2}} \mathrm{~d} x=\mathrm{dt} \\
& \therefore \quad \int \mathrm{e}^{\sin ^{-1} x}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{1-x^2}}\right) \mathrm{d} x=\int \mathrm{e}^{\mathrm{t}}(\sin \mathrm{t}+\cos \mathrm{t}) \mathrm{dt} \\
& =e^t \sin \mathrm{t}+\mathrm{c} \\
& =x \mathrm{e}^{\sin ^{-1} x}+\mathrm{c}
\end{aligned}\)