Evaluate $\int \cos ( \ln x ) d x$
A) $\sin ( \ln x ) + C$

B) $\frac { x ^ { 2 } } { 2 } [ \cos ( \ln x ) + \sin ( \ln x ) ] + C$
C) $\cos \left( \frac { 1 } { x } \right) + C$
D) $\frac { x } { 4 } [ \cos ( \ln x ) + \sin ( \ln x ) ] + C$
E) $\frac { \sin x } { x } + C$
F) $x [ \cos ( \ln x ) + \sin ( \ln x ) ] + C$

G) $- \sin \left( \frac { 1 } { x } \right) + C$

H) $\frac { x } { 2 } [ \cos ( \ln x ) + \sin ( \ln x ) ] + C$

Question

Evaluate $\int \cos ( \ln x ) d x$
A) $\sin ( \ln x ) + C$

B) $\frac { x ^ { 2 } } { 2 } [ \cos ( \ln x ) + \sin ( \ln x ) ] + C$ 
C) $\cos \left( \frac { 1 } { x } \right) + C$ 
D) $\frac { x } { 4 } [ \cos ( \ln x ) + \sin ( \ln x ) ] + C$ 
E) $\frac { \sin x } { x } + C$ 
F) $x [ \cos ( \ln x ) + \sin ( \ln x ) ] + C$

G) $- \sin \left( \frac { 1 } { x } \right) + C$

H) $\frac { x } { 2 } [ \cos ( \ln x ) + \sin ( \ln x ) ] + C$

Quizplus · Accepted Answer

To solve $\int \cos ( \ln x ) d x$, we use integration by parts in the form of the substitution $u = \ln(x)$, which gives $du = \frac{1}{x}dx$, or $dx = x du$. This transforms the integral into $\int \cos(u) x du$. Integrating by parts, we get $x \sin(u) - \int x \sin(u) du$, which simplifies to $x \sin(u) + x \cos(u) + C$ when integrated again by parts. Substituting back $u = \ln(x)$, we get $\frac{x}{2}[\cos(\ln x) + \sin(\ln x)] + C$.

Evaluate ∫cos⁡(ln⁡x)dx\int \cos ( \ln x ) d x∫cos(lnx)dx A) sin⁡(ln⁡x)+C\sin ( \ln x ) + Csin(lnx)+C

Evaluate $\int \cos ( \ln x ) d x$ A) $\sin ( \ln x ) + C$