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Today's calculation of Integral 236

Source:

April 16, 2011
calculusintegrationtrigonometrycalculus computations

Problem Statement

Let aa be a positive constant. Evaluate the following definite integrals A, BA,\ B.
A=∫0πe−axsin⁡2x dx, B=∫0πe−axcos⁡2x dxA=\int_0^{\pi} e^{-ax}\sin ^ 2 x\ dx,\ B=\int_0^{\pi} e^{-ax}\cos ^ 2 x\ dx.
1998 Shinsyu University entrance exam/Textile Science
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