8.pdf - First let uR= sin(5u03b8 dv = e4u03b8 du03b8...

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Unformatted text preview: First let uR= sin(5θ), dv = e4θ dθ ⇒ du =R 5 cos(5θ) dθ, v = 41 e4θ . Then I = e4θ sin(5θ) dθ = 14 e4θ sin(5θ) − 54 e4θ cos(5θ) dθ. dU = −5 sin(5θ) dθ, V = 14 e4θ to NextR let U = cos(5θ), dV = e4θ dθ ⇒ R get e4θ cos(5θ) dθ = 41 e4θ cos(5θ) + 54 e4θ sin(5θ) dθ. Substituting in the previous formulaRgives 5 4θ I = 41 e4θ sin(5θ) − 16 e cos(5θ) − 25 e4θ sin(5θ) dθ 16 1 4θ 5 4θ 25 = 4 e sin(5θ) − 16 e cos(5θ) − 16 I ⇒ 41 5 4θ I = 14 e4θ sin(5θ) − 16 e cos(5θ) + C1 . 16 1 4θ Hence, I= 41 e (4 sin(5θ) − 5 cos(5θ)) + C, where C = 16 C. 41 1 ...
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