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
- A \(\mathrm{e}^{\sin ^{-1} x}+\mathrm{c}\), where c is constant of integration.
- B \(\mathrm{e}^{\sin ^{-1} x} \cdot \sin x+\mathrm{c}\), where c is constant of integration.
- C \(\mathrm{e}^{\sin ^{-1} x} \cdot \cos x+\mathrm{c}\), where c is constant of integration.
- D \(\mathrm{e}^{\sin ^{-1} x} \cdot x+\mathrm{c}\), where c is constant of integration.
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
\(\begin{aligned}
& \text { Put } \sin ^{-1} x=\mathrm{t} \Rightarrow \frac{1}{\sqrt{1-x^2}} \mathrm{~d} x=\mathrm{dt} \\
& \therefore \quad \int \mathrm{e}^{\sin ^{-1} x}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{1-x^2}}\right) \mathrm{d} x=\int \mathrm{e}^{\mathrm{t}}(\sin \mathrm{t}+\cos \mathrm{t}) \mathrm{dt} \\
& =e^t \sin \mathrm{t}+\mathrm{c} \\
& =x \mathrm{e}^{\sin ^{-1} x}+\mathrm{c}
\end{aligned}\)
& \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}\)
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