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8.3 Trigonometric Integrals
∫sin2x dx=?
recall that cos 2x = (1 + cos 2x )/ 2 why?
sin 2x = (1 - cos 2x )/ 2 why?
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how about
∫cos2 2x dx
or ∫ 2sin x cos x dx
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d/dx tan x = sec 2 x, d/dx sec sx = sec x tan x
and tan 2 x + 1 = sec 2x
so applying these identities and derivatives,
∫tan 2x dx= ∫( sec2x - 1 )dx
⇒∫ sec2x dx - ∫dx = tan x - x + c
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November 12, 2015
∫cos3 x sin4 x dx ⇒∫ cos x (cos 2x ) sin4 x dx
⇒∫ cos x (1 - sin 2 x) sin4 x dx
⇒∫ cos x sin4x dx - ∫cos x sin6 x dx
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∫cos3 (x / 3) dx
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∫ (sin5x) dx / (cos x) .5
⇒∫ (1- cos 2x) ( 1- cos2x) cos-.5 sin x dx
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∫sin 4 2θ dθ
⇒∫ sin2 2θ sin2 2θ dθ
⇒∫ ( 1 - cos 2(2 θ)) ( 1- cos 2(2 θ)) / 4 d θ
expand
∫ 1/4 d θ - ∫1/2cos4 θdθ + ∫(1/8 + 1/8 cos 8 θ)dθ
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∫tan 5 (x / 4) dx
answer
sec4 (x /4)-4sec 2 (x / 4) - 4 ln(cos (x /4)) + c
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∫x2 sin2 x dx
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