sin-ln-x-dx-Please-

Question Number 141336 by cesarL last updated on 17/May/21

\int \sin (\ln (x)) d x = ? P l e a s e

Answered by Dwaipayan Shikari last updated on 17/May/21

l o g (x) = u

\int e^{u} s i n (u) d u = Φ

= - e^{u} c o s (u) + \int e^{u} c o s (u) d u

= - e^{u} c o s (u) + e^{u} s i n (u) - \int e^{u} s i n (u) d u

2 Φ = e^{u} (s i n (u) - c o s (u)) Φ = e^{u} (s i n (u) - c o s (u)) + C

= x (s i n (l o g x) - c o s (l o g x)) + C

Answered by qaz last updated on 17/May/21

\int x^{n - 1} d x = \frac{x^{n}}{n} + C

n = 1 + i

x^{i} = e^{i l n x} = \cos (l n x) + i \sin (l n x)

\frac{x^{n}}{n} = \frac{1 - i}{2} [\cos (l n x) + i \sin (l n x)] x

\Rightarrow \int \sin (l n x) d x = \frac{x}{2} [\sin (l n x) - \cos (l n x)] + C

- - - - - - - - - - - - - - - - -

y = l n x

x = e^{y}

\int \sin (l n x) d x = \int e^{y} \sin y d y

= \frac{1}{D} e^{y} \sin y = e^{y} \frac{1}{D + 1} \sin y = e^{y} \frac{1 - D}{1 - D^{2}} \sin y

= e^{y} \frac{1 - D}{1 - i^{2}} \sin y = \frac{e^{y}}{2} (\sin y - \cos y)

= \frac{x}{2} [\sin (l n x) - \cos (l n x)] + C

Commented by cesarL last updated on 17/May/21

I {c o u l d n}^{'} t u n d e r s t a n d

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