cos-log-x-dx-

Question Number 152161 by peter frank last updated on 26/Aug/21

\int \cos (\log x) dx

Answered by Olaf_Thorendsen last updated on 26/Aug/21

F (x) = \int \cos (\log x) d x

F (e^{u}) = \int \cos u e^{u} d u

F (e^{u}) = Re \int e^{i u} e^{u} d u

F (e^{u}) = Re (\frac{e^{i u} e^{u}}{1 + i}) = \frac{1}{2} Re ((1 - i) e^{i u} e^{u})

F (e^{u}) = \frac{1}{2} ((\cos u + \sin u) e^{u})

F (x) = \frac{x}{2} (\cos (\log x) + \sin (\log x)) (+ C)

Commented by peter frank last updated on 26/Aug/21

thank you

Answered by Paradoxical last updated on 26/Aug/21

Commented by peter frank last updated on 26/Aug/21

thank you

Answered by puissant last updated on 26/Aug/21

K = \int c o s (l n x) d x

u = c o s (l n x); v^{'} = 1

\Rightarrow K = x c o s (l n x) + \int s i n (l n x) d x

u = s i n (l n x); v^{'} = 1

\Rightarrow K = x c o s (l n x) + x s i n (l n x) - \int c o s (l n x) d x

\Rightarrow 2 K = x c o s (l n x) + x s i n (l n x)

∴∵ Q = \frac{x}{2} c o s (l n x) + \frac{x}{2} s i n (l n x) + C

Commented by peter frank last updated on 26/Aug/21

thank you

Answered by qaz last updated on 26/Aug/21

\int x^{i} dx = \int e^{ilnx} dx = \frac{x^{i + 1}}{i + 1} + C = \frac{1 - i}{2} x \cdot e^{ilnx} + C

\Rightarrow \int \cos (lnx) dx = \frac{1}{2} xcos (lnx) + \frac{1}{2} xsin (lnx) + C

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