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secondary 4 | A Maths
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Hi,can anyone please explain how to do it.Thank you
Date Posted: 3 years ago
Views: 219
Recall that cos 2A = 2cos²A - 1
So,
2cos²A = 1 + cos 2A
cos² A = ½ + ½ cos 2A
In this question, your A = 3x so 2A = 6x
Therefore,
∫ cos² 3x dx
= ∫ (½ + ½ cos 6x) dx
= ½x + ½(1/6 sin 6x) + c
= ½x + 1/12 sin 6x + c
So,
2cos²A = 1 + cos 2A
cos² A = ½ + ½ cos 2A
In this question, your A = 3x so 2A = 6x
Therefore,
∫ cos² 3x dx
= ∫ (½ + ½ cos 6x) dx
= ½x + ½(1/6 sin 6x) + c
= ½x + 1/12 sin 6x + c
See 1 Answer
done
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clear
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Just apply the addition trigo formula
Date Posted:
3 years ago
① the formula used is actually the reverse of the factor formula (or sum to product formula) and not the addition formula.
Besides, this formula is not covered in O level A Math nowadays.
② integration of cos gives sin and not -sin.
Besides, this formula is not covered in O level A Math nowadays.
② integration of cos gives sin and not -sin.
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