Question

Find the integral
integral of -sec^2 x dx

Answer 1

Super secrit hint!! :O
Do you remember this derivative?\[\Large\rm \frac{d}{dx}\tan x=?\]

Answer 2

yes, sec^2 is the derivative

Answer 3

Good good good.
So the anti-derivative of sec^2x is tanx, yes?

And just carry the negative along for the ride.

Answer 4

but I got -tanx + C as an answer and got it wrong :( ?

Answer 5

Mmmmm you shouldn't have gotten it wrong...
The read the question correctly?
Or is it a definite integral with bounds?

Answer 6

You read*

Answer 7

It just says find the integral, I found the indefinite one for this, do you think it can be done for definite integral?

Answer 8

Well, ya if you're given bounds it can.
Like you might be given limits of pi/4 to 3pi/4 or something like that in radians.
But if you weren't given anything like that, it's fine.

Maybe just a formatting error?
Do you need to put the x in brackets or something?
-tan(x)+C

Answer 9

nope, that is just what I got, let me check again.

Answer 10

No, that's it... maybe its some error in the exercise :/ that's ok, thanks anyways.

Answer 11

weird :U

Answer 12

This is how I would go about it.
To solve
\[\frac {d}{dx}\sec^2x\]
I would use the chain rule where \(g(x)=\sec x\) and \(f(x)=x^2\)
\[\frac{d}{dx}f(g(x))=g'(x)f'(g(x))\]
Popping in \(g(x)\) and \(f(x)\) gives us
\[\frac{d}{dx}f(\sec x ))=(\sec x \tan x )f'(\sec x)\]
\[\frac{d}{dx} \sec^2 x=(\sec x \tan x)(2 \tan x)\]
\[\frac{d}{dx} \sec^2 x= 2 \tan^2 x \sec x\]

Answer 13

Ok, i'll do both ways and see which one would be correct. Thank you!

Answer 14

Hmm that's the derivative, not the integral Saleh.
Maybe that's what the question wanted though heh.

Answer 15

I think that's my eyes telling me to get some sleep!

Answer 16

To sum up what zepdrix mentioned,
\[\frac{d}{dx} \tan x = sec^2x\]
Or
\[d \tan x = \sec^2 x dx\]
Then take a anti-derivative (or integral) from both sides
\[\int d \tan x = \int \sec^2 x dx\]
\[\tan x + c= \int \sec^2 x dx\]
In your case
\[-\int \sec^2 x dx = -\tan x + c \]