∫sin x dx/cos2x
let u = cos x then du = -sin x dx
= -∫du/u2 = 1/u = 1/cos x = sec x
George J.
asked 01/29/20Evaluate the integral
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Question options:
| 1) | 2ln|1 - sin2(x)| + C |
| 2) | -csc(x) + C |
| 3) | sec(x) + C |
| 4) | None of these |
∫sin x dx/cos2x
let u = cos x then du = -sin x dx
= -∫du/u2 = 1/u = 1/cos x = sec x
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