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evaluate: /(sin^2x)/e^x dx?
2 Answers
- Some BodyLv 72 years ago
∫ sin²x / eˣ dx
Start with power reduction formula:
∫ (½ - ½ cos(2x)) / eˣ dx
½ ∫ (1 - cos(2x)) e⁻ˣ dx
Integrate by parts:
u = 1 - cos(2x), du = 2 sin(2x) dx
dv = e⁻ˣ dx, v = -e⁻ˣ
½ [ (1 - cos(2x))(-e⁻ˣ) - ∫ -e⁻ˣ (2 sin(2x) dx) ]
½ [ (cos(2x) - 1) e⁻ˣ + 2 ∫ e⁻ˣ sin(2x) dx ]
½ (cos(2x) - 1) e⁻ˣ + ∫ e⁻ˣ sin(2x) dx
↑ Keep this equation in mind, we'll be coming back to it.
Integrate by parts a second time:
u = sin(2x), du = 2 cos(2x) dx
dv = e⁻ˣ dx, v = -e⁻ˣ
½ (cos(2x) - 1) e⁻ˣ + [ -e⁻ˣ sin(2x) - ∫ -e⁻ˣ (2 cos(2x) dx) ]
½ (cos(2x) - 1) e⁻ˣ - e⁻ˣ sin(2x) + 2 ∫ e⁻ˣ cos(2x) dx
Integrate by parts a third time:
u = cos(2x), du = -2 sin(2x) dx
dv = e⁻ˣ dx, v = -e⁻ˣ
½ (cos(2x) - 1) e⁻ˣ - e⁻ˣ sin(2x) + 2 [ -e⁻ˣ cos(2x) - ∫ -e⁻ˣ (-2 sin(2x) dx) ]
½ (cos(2x) - 1) e⁻ˣ - e⁻ˣ sin(2x) + 2 [ -e⁻ˣ cos(2x) - 2 ∫ e⁻ˣ sin(2x) dx ]
½ (cos(2x) - 1) e⁻ˣ - e⁻ˣ sin(2x) - 2e⁻ˣ cos(2x) - 4 ∫ e⁻ˣ sin(2x) dx
Now remember that this is equal to the equation from earlier:
½ (cos(2x) - 1) e⁻ˣ + ∫ e⁻ˣ sin(2x) dx = ½ (cos(2x) - 1) e⁻ˣ - e⁻ˣ sin(2x) - 2e⁻ˣ cos(2x) - 4 ∫ e⁻ˣ sin(2x) dx
∫ e⁻ˣ sin(2x) dx = -e⁻ˣ sin(2x) - 2e⁻ˣ cos(2x) - 4 ∫ e⁻ˣ sin(2x) dx
5 ∫ e⁻ˣ sin(2x) dx = -e⁻ˣ sin(2x) - 2e⁻ˣ cos(2x)
∫ e⁻ˣ sin(2x) dx = -⅕ e⁻ˣ [ sin(2x) + 2cos(2x) ]
Therefore, substituting into the earlier equation:
½ (cos(2x) - 1) e⁻ˣ - ⅕ e⁻ˣ [ sin(2x) + 2cos(2x) ]
e⁻ˣ [ ½ cos(2x) - ½ - ⅕ sin(2x) - ⅖ cos(2x) ]
e⁻ˣ [ ¹/₁₀ cos(2x) - ½ - ⅕ sin(2x) ]
¹/₁₀ e⁻ˣ [ cos(2x) - 5 - 2 sin(2x) ]
And of course, don't forget the arbitrary constant:
¹/₁₀ e⁻ˣ [ cos(2x) - 5 - 2 sin(2x) ] + C
