Question

e^{2x} sin x x = 6 sin ? dx = 6 cos ? d? x^2 = 36 sin^2 ? ? 1 / sqrt(36 sin^2 ? - 36) 1 / sqrt(36 (sin^2 ? - 1)) 6 sqrt(cos^2 ?) x/6 = 6/6 sin ? sin^-1 (x/6) = sin ? ? (6 cos ? / 6 cos ?) d? ? + C sin^-1 (x/6) + C

          e^{2x} sin x
x = 6 sin ?
dx = 6 cos ? d?
x^2 = 36 sin^2 ?
? 1 / sqrt(36 sin^2 ? - 36)
1 / sqrt(36 (sin^2 ? - 1))
6 sqrt(cos^2 ?)
x/6 = 6/6 sin ?
sin^-1 (x/6) = sin ?
? (6 cos ? / 6 cos ?) d?
? + C
sin^-1 (x/6) + C

Added by Jennifer R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

01/18/2023

Step 1

We start by solving for x: x = -36 Show more…

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e^{2x} sin x x=6 sin θ dx=6cosθ dθ x^2=36 sin^2 θ ∫ 1/sqrt(36 sin^2 θ - 36) dθ x/6 = 6 sin θ / 6 1 / sqrt(36(sin^2 θ - 1)) sin^-1(x/6) = sin θ 6 sqrt(cos^2 θ) ∫ (6 cos θ / 6 cos θ) dθ θ + C sin^-1(x/6) + C