Math, asked by priyanshi29108, 1 year ago

integration of sin(log x)dx

Answers

Answered by harshbhardwaj23
1
∫ 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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Answered by iHelper
4
Hello!

\int \sf sin(\sf log\:x)\: \sf dx

\int \sf e^{\sf log\:x} \:sin(\sf log\:x)\: \sf \:d(\sf log\:x)

\sf e^{\sf y} \int \sf sin(\sf y) \sf dy - \int \dfrac{\sf d}{\sf dy}(\sf e^{\sf y}) \int \sf sin(\sf y) \sf dy

 - \sf e^{\sf y}cos\:y + \sf e^{\sf y}sin\:y - \sf I + \sf C

\dfrac{\sf e^{\sf y}}{\sf 2}(\sf sin\:y - cos\:y)

\dfrac{\sf x}{\sf 2} (\sf sin(log\:x) - \sf cos(log\:x)) + \sf C

Cheers!
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