MHT CET · Maths · Indefinite Integration
\(\int \frac{e^{\frac{x}{2}}}{\sqrt{e^{-x}-e^x}} \mathrm{~d} x=\sin ^{-1}(f(x))+C, \quad\) (where \(\quad C \quad\) is constant of integration.) then \(f(2)\) has the value
- A \(e\)
- B \(e^2\)
- C \(e^{\frac{1}{2}}\)
- D \(e^{\frac{3}{2}}\)
Answer & Solution
Correct Answer
(B) \(e^2\)
Step-by-step Solution
Detailed explanation
\(\int \frac{e^{x / 2}}{\sqrt{e^{-x}-e^x}} \mathrm{~d} x=\int \frac{e^x \mathrm{~d} x}{\sqrt{1-\left(e^x\right)^2}}[\text { Multiplying }\) \(N^r \text { and } D^r \text { by } e^{\frac{x}{2}} \text { ] }\)
\( =\int \frac{\mathrm{d} t}{\sqrt{1-t^2}}=\sin ^{-1} t+C=\sin ^{-1}\left(e^x\right)+C\left[\text { let } e^x=t\right] \)
\( \Rightarrow f(x)=e^x \)
\( \Rightarrow f(2)=e^2\)
\( =\int \frac{\mathrm{d} t}{\sqrt{1-t^2}}=\sin ^{-1} t+C=\sin ^{-1}\left(e^x\right)+C\left[\text { let } e^x=t\right] \)
\( \Rightarrow f(x)=e^x \)
\( \Rightarrow f(2)=e^2\)
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