Question

integration of sin(log x)dx

Answer 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

Ｉｆ ｙｏｕ ｌｉｋｅ ａｎｓｗｅｒ ａｄｄ ｔｏ ＢＲＡＩＮＬＩＳＴ.

Answer 2

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!