∫
x
cos
(
2
x
)
d
x
\int x\cos{(2x)} dx
∫
x
cos
(
2
x
)
d
x
=
1
2
x
2
sin
(
2
x
)
+
C
\displaystyle \frac{1}{2}x^2\sin{(2x)}+C
2
1
x
2
sin
(
2
x
)
+
C
1
2
x
2
cos
(
2
x
)
+
1
2
sin
(
2
x
)
+
C
\displaystyle \frac{1}{2}x^2\cos{(2x)}+\frac{1}{2}\sin{(2x)}+C
2
1
x
2
cos
(
2
x
)
+
2
1
sin
(
2
x
)
+
C
1
2
x
sin
(
2
x
)
−
1
4
cos
(
2
x
)
+
C
\displaystyle \frac{1}{2}x\sin{(2x)}-\frac{1}{4}\cos{(2x)}+C
2
1
x
sin
(
2
x
)
−
4
1
cos
(
2
x
)
+
C
1
2
x
sin
(
2
x
)
+
1
4
cos
(
2
x
)
+
C
\displaystyle \frac{1}{2}x\sin{(2x)}+\frac{1}{4}\cos{(2x)}+C
2
1
x
sin
(
2
x
)
+
4
1
cos
(
2
x
)
+
C
Summary
Submit
Skip Question
Approach
Use integration by parts:
u
=
x
d
v
=
cos
(
2
x
)
d
x
u= x \hskip{2em} dv= \cos{(2x)} dx
u
=
x
d
v
=
cos
(
2
x
)
d
x
d
u
=
d
x
v
=
1
2
sin
(
2
x
)
du=dx \hskip{2em} v= \frac{1}{2}\sin{(2x)}
d
u
=
d
x
v
=
2
1
sin
(
2
x
)
∫
u
d
v
=
u
v
−
∫
v
d
u
\int u dv = uv - \int v du
∫
u
d
v
=
uv
−
∫
v
d
u
∫
x
cos
(
2
x
)
d
x
=
x
(
1
2
sin
(
2
x
)
)
−
∫
1
2
sin
(
2
x
)
d
x
\int x\cos{(2x)} dx = x\left(\frac{1}{2}\sin{(2x)}\right) - \int \frac{1}{2}\sin{(2x)} dx
∫
x
cos
(
2
x
)
d
x
=
x
(
2
1
sin
(
2
x
)
)
−
∫
2
1
sin
(
2
x
)
d
x
=
1
2
x
sin
(
2
x
)
+
1
4
cos
(
2
x
)
+
C
= \boxed{\frac{1}{2}x\sin{(2x)}+\frac{1}{4}\cos{(2x)}+C }
=
2
1
x
sin
(
2
x
)
+
4
1
cos
(
2
x
)
+
C