integration of sin2xcos3x
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∫ sin2x cos3x dx in this question following formula will be used
2sinA cosB=sin(A+B)+sin(A-B)
ans- 1/2∫ 2sin2x cos3x dx
=1/2∫ sin(2x+3x)+sin(2x-3x) dx
=1/2∫ sin5x+sin(-x) dx { sin(-θ )= -sinθ }
=1/2∫ sin5x- sinx dx
=1/2∫ sin5x dx -1/2 ∫ sinx dx
=1/10(-cos5x)+1/2cosx (by integration)
2sinA cosB=sin(A+B)+sin(A-B)
ans- 1/2∫ 2sin2x cos3x dx
=1/2∫ sin(2x+3x)+sin(2x-3x) dx
=1/2∫ sin5x+sin(-x) dx { sin(-θ )= -sinθ }
=1/2∫ sin5x- sinx dx
=1/2∫ sin5x dx -1/2 ∫ sinx dx
=1/10(-cos5x)+1/2cosx (by integration)
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