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Mathematics
Problem
Which one of the following gives the correct antiderivative of
e
a
x
(
sin
b
x
)
e^{ax}\bigl(\sin bx\bigr)
e
a
x
(
sin
b
x
)
, where
a
,
b
∈
R
a,\,b\in\mathbb{R}
a
,
b
∈
R
and
a
2
+
b
2
≠
0
a^{2}+b^{2}\neq0
a
2
+
b
2
=
0
? a)
e
a
x
(
a
cos
b
x
−
b
sin
b
x
)
a
2
+
b
2
+
C
\displaystyle \frac{e^{ax}\bigl(a\cos bx-b\sin bx\bigr)}{a^{2}+b^{2}}+C
a
2
+
b
2
e
a
x
(
a
cos
b
x
−
b
sin
b
x
)
+
C
b)
e
a
x
(
a
sin
b
x
−
b
cos
b
x
)
a
2
+
b
2
+
C
\displaystyle \frac{e^{ax}\bigl(a\sin bx-b\cos bx\bigr)}{a^{2}+b^{2}}+C
a
2
+
b
2
e
a
x
(
a
sin
b
x
−
b
cos
b
x
)
+
C
c)
e
a
x
(
b
sin
b
x
−
a
cos
b
x
)
a
2
+
b
2
+
C
\displaystyle \frac{e^{ax}\bigl(b\sin bx-a\cos bx\bigr)}{a^{2}+b^{2}}+C
a
2
+
b
2
e
a
x
(
b
sin
b
x
−
a
cos
b
x
)
+
C
d)
e
a
x
(
a
sin
b
x
−
b
cos
b
x
)
a
2
−
b
2
+
C
\displaystyle \frac{e^{ax}\bigl(a\sin bx-b\cos bx\bigr)}{a^{2}-b^{2}}+C
a
2
−
b
2
e
a
x
(
a
sin
b
x
−
b
cos
b
x
)
+
C
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