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AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

Section: Mathematics

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Question : 76 of 160

Marks: +1, -0

\int\limits_{\log_e 2}^{x}

\frac{dt}{\sqrt{e^t-1}}

\frac{\pi}{6}

⇒ x =

2 . $\log_e$ 2
3 . $\log_e$ 2
4 . $\log_e$ 2
8 . $\log_e$ 2

Solution:

Consider the given integral,

\int\limits_{\log_{e} 2}^{x} \;\frac{dt}{\sqrt{e^{t}-1}}=\;\frac{\pi}{6}

\int\limits_{\log_{e} 2}^{x} \;\frac{e^{-\frac{t}{2}}}{\sqrt{1-\left(e^{-\frac{t}{2}}\right)^2}} dt=\;\frac{\pi}{6}

Take

e^{-\frac{t}{2}}=u

then,

e^{-\frac{t}{2}} dt=-2 du

\text{At } t=\log_{e} 2, u=\;\frac{1}{\sqrt{2}}

\;\text{At }\; t=x, u=e^{-\frac{x}{2}}

Substitute these values in the above expression.

\int\limits_{\frac{1}{\sqrt{2}}}^{e^{\frac{x}{2}}} \frac{-2 du}{\sqrt{1-u^{2}}} \;\; =\frac{\pi}{6}

\sin^{-1}\left(e^{-\frac{x}{2}}\right)-\frac{\pi}{4} \;\; =-\frac{\pi}{12}

\sin^{-1}\left(e^{-\frac{x}{2}}\right) \;\; =\frac{\pi}{6}

e^{-\frac{x}{2}} \;\; =\frac{1}{2}

-\frac{x}{2} \;\; =\log_{e}\left(\frac{1}{2}\right)

\;\; =-\log_{e}(2)

x \;\; =2\log_{e}(2)

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