Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_0^{2 \pi} e ^{\frac{x}{2}} \cdot \sin \left(\frac{x}{2}+\frac{\pi}{4}\right) d x=$
A
1
B
$2 \sqrt{2}$
✓
$0$
D
None of these
✓
Answer
Correct option: C.
$0$
(C) Let $I =\int_0^{2 \pi} e ^{\frac{x}{2}} \cdot \sin \left(\frac{x}{2}+\frac{\pi}{4}\right) d x$ Put $\frac{x}{2}= t$ $\Rightarrow d x=2 dt$ $\therefore \quad I=2 \int_0^\pi e^t \sin \left(t+\frac{\pi}{4}\right) d t$ $=2\left[\frac{ e ^{ t }}{\sqrt{1+1}} \sin \left( t +\frac{\pi}{4}-\tan ^{-1} \frac{1}{1}\right)\right]_0^\pi$ $\left[\because \int e ^{ ax } \sin b x d x=\frac{ e ^{ ax }}{\sqrt{ a ^2+ b ^2}} \sin \left(b x-\tan ^{-1} \frac{b}{ a }\right)+ c \right]$ $\begin{array}{l}=\frac{2}{\sqrt{2}}\left[ e ^{ t } \sin t\right]_0^\pi \\ =\frac{2}{\sqrt{2}}[0]=0\end{array}$
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