Integral of sec (y) dy?

Question

Why can't we find the integral by using the defintion

sec(y) = 1/cos(y) ?? and the answer would be ln l (cos(y)) l +c

the tutorial i'm solving the problem from states that we can't do that and we should multiply the top and bottom of the fraction by a suitable term ( secy + tan y in this case) .. and the answer turned out to be >>

ln l sec y + tan y l + c

Answer 1

simple concept to follow here

Suppose we have x = ln | cos(y)| + c,

when we differentiate x with respect to y,
applying Chain rule here,

dx/dy = (1/cos(y))(-sin(y)) = -tan (y) and not sec(y).

But when we differentiate x = ln | sec(y) +tan(y) | +c with respect to y,
again using Chain rule,

(d/dy)(x) = (1/(sec(y) + tan (y))).(sec(y).tan(y) + sec(^2)(y))
dx/dy = (1/(sec(y) + tan(y))).sec(y).(tan(y) + sec(y))

Since we cancel (1/(sec(y) + tan(y))) & (sec(y) + tan(y)),
Hence we get dx/dy = sec(y).

As we know that integration is nothing but reverse of differentiation, where we integrate the differential result to get the value we differentiated to get this result.

Hence we don't convert sec(y) to (1/cos(y)).

Answer 2

When you write that integral
(1/y dy) = ln|y| +c
you are right, and the reason of this is the fact that the variable of integrating (y in this case) is the same as the variable in the expression 1/y. So, in your case you would get ln|cos(y)| only if consider integral (1/cos(y) d cos(y)). And it is not the same. If while integrating (1/cos(y) d y) multiply the top and the bottom of the fraction by (sec y + tan y) you will have:
integral( 1/cos^2(y) + sin(y)/cos^2(y) )/(sec y + tan y) dy =
integral 1/(sec y + tan y) d(sec y + tan y) = ln |sec y + tan y| + c

Answer 3

For the sake of argument, let's assume that the integral is indeed ln|cos(y)| + C

If f(y) = ln|cos(y)| + C and we differentiate, we should, by your logic, get sec(y). However, that's not the case. Applying the chain rule,

f'(y) = 1/cos(y) [-sin(y)]
f'(y) = -sin(y)/cos(y) = -tan(y)

So you've unintentionally discovered the integral of -tan(y) when you stated ln| cos(y) | + C

Now, let's take the derivative of f(y) = ln |sec(y) + tan(y)| + C.

f(y) = ln |sec(y) + tan(y)| + C

Applying the chain rule,

f'(y) = 1/[sec(y) + tan(y)] { sec(y)tan(y) + sec^2(y) }

Factor sec(y),

f'(y) = 1/[sec(y) + tan(y)] { sec(y) [tan(y) + sec(y)] }

Note the cancellation. This leaves us with just

f'(y) = sec(y)

Answer 4

Let us consider the tutorial suggestion:-
I = ∫sec y.dy
I = ∫ secy.(sec y + tan y) / (sec y + tan y).dy
Let u = sec y + tan y
du = (sec y.tan y + sec²y).dy
du = sec y.(tan y + sec y).dy
I = ∫(1 / u) du
I = log u + C
I = log (secy + tan y) + C

Answer 5

combine through aspects. allow u = y du = dy dv = sec²y dy v = tan y ? y sec²y dy = y tan y - ? tan y dy ? tan y dy = ? (sin y/cos y) dy allow z = cos y Then dz = -sin y dy ? tan y dy = -? dz/z = -ln|z| = -ln |cos y| for this reason, ? y sec²y dy = y tan y + ln |cos y| + C

Answer 6

Integral of dx/x = ln|x| + c

So integral d(sec x)/sec x = ln |sec x| +c

Answer 7

...

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