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618

Part of 2010 Today's Calculation Of Integral

Problems(1)

Today's calculation of Integral 618

Source:

7/1/2010
Find the minimu value of 1π∫−π2π2{xcos⁡t+(1−x)sin⁡t}2dt.\frac{1}{\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \{x\cos t+(1-x)\sin t\}^2dt.
2010 Ibaraki University entrance exam/Science
calculusintegrationtrigonometrycalculus computations