25 ++ π/2 9 cos2(θ) dθ 0 238305-π/2 9 cos 2(θ) dθ 0

4/22/18 · Issuu is a digital publishing platform that makes it simple to publish magazines, catalogs, newspapers, books, and more online Easily share your publications and getVeja grátis o arquivo Howard, Anton Calculo Um Novo Horizonte Exercícios resolvidos capitulo 15 enviado para a disciplina de Cálculo III Categoria Outro 2For ∫cos2 3θdθ For a correct integral expression including limits (may be implied 6 9 14 0 (1 )( 7 ) 2(2 4) 3(2 2) 0 3 2 2 Correct shape for π θ π 2 1 2 −1 < < including maximum in 1st quadrant Correct form at O and no extra sections (ii) 3 Area is

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π/2 9 cos 2(θ) dθ 0

π/2 9 cos 2(θ) dθ 0-Anuncio 3 r2 = 9 sin 2θ, r ≥ 0, 0 ≤ θ ≤ π/2 ∫ π 2 1 · 9 sin 2θdθ Area = 0 2 π 2 9 = − cos 2θ 4 0 9 = 2 4 r = tan θ, π/6 ≤ θ ≤ π/3 ∫ π 3 1 Area = tan2 θdθ π 2 6∫ π 1 3 = sec2 θ − 1dθ 2 π6 1 √ 1 π π = ( 3− √ − ) 2 6 3 3 1√ 1 3− π = 3 12 11 r = 3 2 cos θ ∫ 1 π Area = 2 · (3 2 cos θ)2 dθ 2 0 ∫ π = 9 12 cos θ 4 cos2 θdθ 0 = 11π 23 ∫ π 3µ 2 k " 1/ 2 $ kτ 4 π2µ % 1/2!π 4 " = τ 4 √ 2 as advertised Problem 36 (a) Show that if a particle describes a circular orbit under the influence of an attractive central force directed at a point on the circle, then the force varies as the

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$\begingroup$ $9\sin^2 \theta1=0$ gives $\sin \theta=\pm \frac13$ so there are in fact 3 solutions $\endgroup$ – rae306 Aug 2 '15 at $\begingroup$ The sine is $\pm \frac{1}{3}$ The domain of the trig functions has not been specified, so there are infinitely many answers, tough to do a commaseparated list $\endgroup$ – AndréRπ/4 0 35cos2θ cos2θ dθ = Rπ/4 0 3 cos2θ 5cos2θ cos2θ dθ = Rπ/4 0 (3sec2θ 5)dθ =3tanθ 5θπ/4 0 = 3tanπ 4 5π 4 −(00)=35π 4I Limits in θ θ ∈ 0π;

Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history7/24/14 · Solución Por simetría, se calculará 4 veces el área de la porción en el primer cuadrante cos 2θ = 0 ⇒ θ = π 4 , en el primer cuadrante A = 4 ⋅ ∫0 π 4 1 2 r2 dθ = 2∫0 π 4 r2 dθ = 2∫0 π 4 9cos 2θ dθ A = 18∫0 π 4 cos 2θ dθ = 9 Área en coordenadas polares 5Evaluate The Integral π/2 3 Cos2θ Dθ 0 ;

I Limits in φ φ ∈ 0,π/2;4/13/19 · 0 # 0 π/2 sin2 xdx =!(c) Determine whether the imroper integral R ?

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Sin3 x cos xdx =?5/31/10 · Writeacher May 31, 10 When the sin/cos are in odd powers, use the substitution cosθdθ=d (sinθ), or sinθdθ=d (cosθ) ∫sin³θcos³θd&theta (from 0 to π/2) =∫sin³θ (1sin²θ) (cosθdθ) =∫ (sin³θsin 5 θ) (d (sinθ)) (from 0 to sin (π/2)) = sin 4 θ/4sin 6 θ/6 (from 0 to 1)11/3/12 · ASTM002 Galaxies Lecture notes 06 (QMUL),

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See the answer Evaluate the integral π/2 3 cos 2 θ dθ 0 Videos Stepbystep answer 0150 0 0 Expert Answer 100% (63 ratings) Previous question Nextπ/2 cosn (x)dx = 1 n−1( x)sin( π/2 n − 1 π/2 cosn−1(x )dx 0 n 0 n 0 Observe nthat cos −1(x) sin(x)π/2 is zero for n ≥ 2 Let us call 0 π/2 A n = ncos (x)dx 0 Combined with this observation, the recursion formula of (i) gives A n = n − 1 A n−2, n ≥ 2 n = 1 n = 1 (6) = π/2 n = 0 Here the values have been found directly for n = 0 and n = 1π 2 −π 2 dθ acosθ 0 rdr= π 2 −π 2 1 2 r2 acosθ 0 dθ = a2 2 π 2 −π 2 cos2 θdθ= a2 2 π 2 −π 2 cos2θ1 2 dθ = π 4 a2 (2) 積分の順序を変更して計算すると, 2 1 dx 2 1 x yexy dy = 1 1 2 dy 2 1 y yexy dx 2 1 dy 2 1 yexy dx = 1 1 2 exy 2 1 y dy 2 1 exy 2 dy = 1 1 2 (e2y − e) 1 = ( 解答

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Evaluate the integralπ/47 9 cos2 (θ)cos2 (θ) dθ Evaluate the integralπ/47 9 cos2 (θ)cos2 (θ) dθ Watch laterUse a graphing utility to graph three solutions, one of which passes through the given point ds/dθ = tan 2θ, (0, 2)3 4), (3 4,´ 3?

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(b) Evaluate the integral R 2x 2x2?5x?3 =?(1 − cos2 = 9 2 θ− 1 2 sen 2θ π/2 π/4 = 9 8 (π 2) 8 A = 3 π/2 π/2 1 2 θ 2 dθ = 1 10 θ5 3 /2 /2 = 121 160 π5 9 4 A = π 0 1 2 (5sen θ)2 dθ = 25 4 π 0 (1 − cos2 θ) dθ = 25θ− 1 2 sen 2θ π 0 = 4 π 10 2 A = 2 π/2 −π/2 1 2 (4 − senθ) 2 dθ = π/2 −π/2 16 − 8senθ sen2 θ dθ = π/2 −π/2 16 sen2 θ dθ pelo Teorema 557(b) = 2 π/2 0 16 sen2 θ dθ pelo Teorema 557(a) = 2 π/2 0 16 1 2 (1 − cos2 θ) dθ0 cos2 θ dθ = 0 1 2 (1 cos 2θ) dθ halfangle identity 1 £ 1 ¤π/2 £¡ ¢ ¤ = 2 θ 2 sin 2θ 0 = 12 π2 0 − (0 0) = π 4 Rπ Rπ£ ¤2 Rπ£ ¤2 Rπ 9

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