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Integrate the following functions: sin 3x cos4x |
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Answer» SOLUTION :`sin 3X cos 4x = cos 4x sin 3x =1/2[sin(4x+3x)-sin(4x-3x)] =1/2[sin7x-sinx] therefore `int SIN3X cos4x dx` =1/2[-cos(7X)/7+cosx]+C |
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