Find the integrals of the function:
kanak001:
sorry
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wait for awhile ..here u are ..miss
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we have to find ∫cos⁴(2x) dx
first of all resolve the cos⁴(2x) into simpler form.
we know, cos²Φ = (1 + cos2Φ)/2
so, cos⁴(2x) = [cos²(2x) ]²
= [{1 + cos2(2x)}/2 ]²
= [(1 + cos4x)/2 ]²
= (1 + cos4x)²/4
= {1 + cos²(4x) + 2cos(4x)}/4
= 1/4 + cos²(4x)/4 + cos(4x)/2
= 1/4 + {1 + cos2(4x)}/8 + cos(4x)/2
= 1/4 + 1/8 + cos(8x)/8 + cos(4x)/2
= 3/8 + cos(8x)/8 + cos(4x)/2
now, ∫cos⁴(2x) dx = ∫[3/8 + cos(8x)/8 + cos(4x)/2 ] dx
= 3/8∫dx + 1/8∫cos(8x) dx + 1/2∫ cos(4x) dx
= 3x/8 + 1/8 sin(8x)/8 + 1/2 sin(4x)/4 + K, where K is constant
= 3x/8 + 1/64 sin(8x) + 1/8 sin(4x) + K,
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