integration of sin(log x)dx
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∫ sin(ln(x)) dx [1 , e] => use integration by parts:
u = sin(ln x) , dv = dx
du = cos(ln x) * 1/x dx , v = x
∫ sin(ln (x)) dx = x sin(ln (x)) - ∫ cos(ln (x)) dx => 2nd integration by parts:
u = cos(ln(x)) , dv= dx
du = -sin(ln(x)) * 1/x dx , v = x
= x sin(ln (x)) - x cos(ln(x)) - ∫ sin(ln(x)) dx
= ½ * x * [sin(ln(x)) - cos(ln(x))] [1 , e]
= ½ [e sin(1) - e cos(1) + 1] ≈ 0.909
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u = sin(ln x) , dv = dx
du = cos(ln x) * 1/x dx , v = x
∫ sin(ln (x)) dx = x sin(ln (x)) - ∫ cos(ln (x)) dx => 2nd integration by parts:
u = cos(ln(x)) , dv= dx
du = -sin(ln(x)) * 1/x dx , v = x
= x sin(ln (x)) - x cos(ln(x)) - ∫ sin(ln(x)) dx
= ½ * x * [sin(ln(x)) - cos(ln(x))] [1 , e]
= ½ [e sin(1) - e cos(1) + 1] ≈ 0.909
If you like answer add to BRAINLIST.
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