Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
If $\int_{\log 2}^x \frac{ du }{\left( e ^{ u }-1\right)^{1 / 2}}=\frac{\pi}{6}$, then $e ^x=$
A
1
B
2
✓
4
D
-1
✓
Answer
Correct option: C.
4
(C) $\int_{\log 2}^x \frac{ du }{\left( e ^{ u }-1\right)^{1 / 2}}=\frac{\pi}{6} \Rightarrow \int_{\log 2}^x \frac{ e ^{ u }}{ e ^{ u }\left( e ^{ u }-1\right)^{1 / 2}} du =\frac{\pi}{6}$ Put $e^u-1=t^2 \Rightarrow e^u d u=2 t d t$ When $u =\log 2, t =1$ and when $u =x, t =\sqrt{ e ^x-1}$ $\therefore \quad \int_1^{\sqrt{ e ^x-1}} \frac{2}{1+ t ^2} dt =\frac{\pi}{6}$ $\Rightarrow 2\left[\tan ^{-1} t \right]_1^{\sqrt{ e ^x-1}}=\frac{\pi}{6}$ $\Rightarrow \tan ^{-1}\left(\sqrt{ e ^x-1}\right)-\frac{\pi}{4}=\frac{\pi}{12}$ $\Rightarrow \sqrt{ e ^x-1}=\tan \frac{\pi}{3} \Rightarrow \sqrt{ e ^x-1}=\sqrt{3} \Rightarrow e ^x=4$
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