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2007-10-09 11:03:20 · 4 answers · asked by Integer 1 in Science & Mathematics Mathematics

4 answers

∫cos^2(x) dx
= ∫(1/2)(cos(2x)+1) dx
= (1/4)sin(2x) + (1/2)x + c

2007-10-09 11:11:15 · answer #1 · answered by sahsjing 7 · 0 0

You need the following trig identity:
(cos x)^2 = 1/2(1+cos 2x)
Then,
∫ (cos x)^2 dx
= 1/2 ∫ 1+ cos 2x dx
= x/2 + 1/2 ∫ cos 2x dx
= x/2 + 1/2 (sin 2x)/2
= x/2 + (sin 2x)/4

2007-10-09 11:13:34 · answer #2 · answered by 7 · 0 0

I = ∫ cos ² x dx
I = ∫ (1/2) cos 2x + 1/2 dx
I = (1/4) sin 2x + (1/2) x + C

2007-10-09 11:20:58 · answer #3 · answered by Como 7 · 1 0

Integral of (cosx)^2 = [(cosx)^3]/[-3sinx]

2007-10-09 11:19:14 · answer #4 · answered by Peter 1 · 0 0

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