Evaluate the integral.
-$$\int 15 x \cos \frac { 1 } { 2 } x d x$$
A) $30 x \sin \left( \frac { 1 } { 2 } \right) x + 60 \cos \left( \frac { 1 } { 2 } \right) x + C$
B) $60 \sin \left( \frac { 1 } { 2 } \right) x - 30 x \cos \left( \frac { 1 } { 2 } \right) x + C$
C) $15 x \sin \left( \frac { 1 } { 2 } \right) x - 30 \cos \left( \frac { 1 } { 2 } \right) x + C$
D) $15 \sin \left( \frac { 1 } { 2 } \right) x + 30 x \cos \left( \frac { 1 } { 2 } \right) x + C$

Question

Quizplus · Accepted Answer

Using integration by parts with $u = 15x$ and $dv = \cos \frac{1}{2}x dx$, we get $du = 15 dx$ and $v = 2 \sin \frac{1}{2}x$. So,$$\int 15 x \cos \frac { 1 } { 2 } x d x = 15x \sin \frac { 1 } { 2 } x - \int 15 \sin \frac { 1 } { 2 } x dx$$$$= 15x \sin \frac { 1 } { 2 } x - 30 \cos \frac { 1 } { 2 } x + C$$

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Evaluate the Integral $\int 15 x \cos \frac { 1 } { 2 } x d x$