Jawaban:
Gunakan integral parsial.
Diketahui bahwa ∫u dv = uv - ∫v du
Misalkan:
u = e^x, maka du = e^x dx
dv = sin(x) dx, maka v = -cos(x)
Jadi,
∫(e^x * sin(x)) dx = (e^x)(-cos(x)) - ∫(-cos(x))(e^x dx)
= -e^x(cos(x)) + ∫ e^x * cos(x) dx
dv = cos(x) dx, maka v = sin(x)
maka,
-e^x(cos(x)) + ∫ e^x * cos(x) dx = -e^x(cos(x)) + [(e^x)(sin(x) - ∫(sin(x))(e^x dx)]
= e^x(sin(x)) - e^x(cos(x)) - ∫(e^x * sin(x)) dx
= e^x(sin(x) - cos(x)) - ∫(e^x * sin(x)) dx
Jadi, akan kita dapatkan
∫(e^x * sin(x)) dx = e^x(sin(x) - cos(x)) - ∫(e^x * sin(x)) dx
∫(e^x * sin(x)) dx + ∫(e^x * sin(x)) dx = e^x(sin(x) - cos(x))
2∫(e^x * sin(x)) dx = e^x(sin(x) - cos(x))
∫(e^x * sin(x)) dx = (e^x(sin(x) - cos(x)))/2 + C
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Jawaban:
Gunakan integral parsial.
Diketahui bahwa ∫u dv = uv - ∫v du
Misalkan:
u = e^x, maka du = e^x dx
dv = sin(x) dx, maka v = -cos(x)
Jadi,
∫(e^x * sin(x)) dx = (e^x)(-cos(x)) - ∫(-cos(x))(e^x dx)
= -e^x(cos(x)) + ∫ e^x * cos(x) dx
Misalkan:
u = e^x, maka du = e^x dx
dv = cos(x) dx, maka v = sin(x)
maka,
-e^x(cos(x)) + ∫ e^x * cos(x) dx = -e^x(cos(x)) + [(e^x)(sin(x) - ∫(sin(x))(e^x dx)]
= e^x(sin(x)) - e^x(cos(x)) - ∫(e^x * sin(x)) dx
= e^x(sin(x) - cos(x)) - ∫(e^x * sin(x)) dx
Jadi, akan kita dapatkan
∫(e^x * sin(x)) dx = e^x(sin(x) - cos(x)) - ∫(e^x * sin(x)) dx
∫(e^x * sin(x)) dx + ∫(e^x * sin(x)) dx = e^x(sin(x) - cos(x))
2∫(e^x * sin(x)) dx = e^x(sin(x) - cos(x))
∫(e^x * sin(x)) dx = (e^x(sin(x) - cos(x)))/2 + C