∫logπ/2logπe2xsec213e2xdx is equal to

I=∫logπ/2logπe2xsec213e2xdx Put e2x=t⇒2e2xdx=dt When x=logπ/2,t=e2logπ/2=elogπ/2=π2 When x=logπ,t=e2logπ=πI=∫π2π12sec213tdt=12⋅113tant3⋅π/2π=32tanπ3−tanπ6=323−13=3

∫logπ/2logπe2xsec213e2xdx is equal to

I=∫logπ/2logπe2xsec213e2xdx Put e2x=t⇒2e2xdx=dt When x=logπ/2,t=e2logπ/2=elogπ/2=π2 When x=logπ,t=e2logπ=πI=∫π2π12sec213tdt=12⋅113tant3⋅π/2π=32tanπ3−tanπ6=323−13=3

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$\int_{\log \sqrt{π / 2}}^{\log \sqrt{π}} e^{2 x} \sec^{2} (\frac{1}{3} e^{2 x}) dx is equal to$

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$\frac{1}{\sqrt{3}}$

$\sqrt{3}$

$\frac{1}{2 \sqrt{3}}$

$\frac{3 \sqrt{3}}{2}$

answer is A.

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Detailed Solution

$\begin{matrix} I = \int_{\log \sqrt{π / 2}}^{\log \sqrt{π}} e^{2 x} \sec^{2} (\frac{1}{3} e^{2 x}) dx \\ {Put e}^{2 x} = t \Rightarrow 2 e^{2 x} dx = dt \\ When x = \log \sqrt{π / 2}, t = e^{2 \log \sqrt{π / 2}} = e^{\log π / 2} = \frac{π}{2} \\ When x = \log \sqrt{π}, t = e^{2 \log \sqrt{π}} = π \\ I = \int_{\frac{π}{2}}^{π} \frac{1}{2} \sec^{2} (\frac{1}{3} t) d t \\ = \frac{1}{2} \cdot \frac{1}{\frac{1}{3}} {[\tan \frac{t}{3} \cdot]}_{π / 2}^{π} \\ = \frac{3}{2} [\tan \frac{π}{3} - \tan \frac{π}{6}] \\ = \frac{3}{2} [\sqrt{3} - \frac{1}{\sqrt{3}}] \\ = \sqrt{3} \end{matrix}$