InterviewSolution
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InterviewSolution
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`int_(-pi//2)^(pi//2)sin^(2)xdx`A. `pi/2`B. `pi/4`C. `pi/2-1/4`D. `pi/4-1/2` |
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Answer» Correct Answer - A Let I = \(\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\) \(\because\) sin2(-x) = (-sin x)2 = sin2x \(\therefore\) I = 2 \(\int\limits^{\frac{\pi}2}_0\) sin2x dx = \(\cfrac{2\,\Gamma(\frac{2+1}{2})\,\Gamma(\frac{0+1}{2})}{2\,\Gamma\,(\frac{2+0+2}{2})}\) = \(\cfrac{\Gamma(\frac32)\,\Gamma(\frac12)}{\Gamma(2)}\) = \(\cfrac{\frac12\,\Gamma(\frac12)\,\Gamma(\frac12)}{1!}\) (\(\because\Gamma(\frac32) = \frac12\Gamma(\frac12)\)) = \(\frac12\sqrt{\pi}.\sqrt{\pi}\) (\(\because\Gamma(\frac12)= \sqrt{\pi}\)) = \(\frac{\pi}2\) |
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