A slightly interesting and simple integral, can anyone tell me why this is true?

Question

kainui · Answer

$$\int\limits_{}^{} \tan (x) \sec^2 (x) dx $$ So if you pick u=sec(x) you get a different answer than if you pick u=tanx. Why is that?

dls · Answer

The change is adjusted by the integration constant C which you apply normally I guess,if you take some interval and evaluate it as a definite integral you will get the same answer no matter what term you let=u.

kainui · Answer

Yeah, I guess I was looking more for the fact that:

tan^2x+1=sec^2x so taking the derivative of both sides will give you the same thing.

ganeshie8 · Answer

1/2sec^2x + c1 =  1/2tan^2x + c2

ganeshie8 · Answer

what DLS said is also same...

ganeshie8 · Answer

1 is absorbed by C

kainui · Answer

Out of curiosity when you see that integral what does everyone see first the tangent or secant integral?

ganeshie8 · Answer

myself u = tanx ofcourse :)

u = secx dint occur to me till date... it kept me thinkning few seconds.. when u asked to use u = secx

anonymous · Answer

Yes DLS is right

if you do the first method

you get answer as $$I = \frac{(tanx)^2}{2} = \frac{\sec^2x}{2}-\frac{1}{2}$$

kainui · Answer

Yeah, I'm just looking for clever/interesting integrals. It is sort of interesting to think that tan^2x has the same derivative as sec^2x, I suppose when i think of them they seem like they should be much more different from each other, but really one is just shifted up by 1 of the other haha.

ganeshie8 · Answer

lol u gona become an evil professor someday :P

kainui · Answer

Good. Err... Bad. =)

anonymous · Answer

http://www.wolframalpha.com/input/?i=tan%5E2%28x%29 http://www.wolframalpha.com/input/?i=sec%5E2%28x%29 but they have the same graph :D.. no wonder they have the same slope.. !!

kainui · Answer

I'm noticing something interesting here, this implies an infinite definition...$$\sec^2x=1+\sin^2x \sec^2x$$$$\sec^2x=1+\sin^2x (1+\sin^2x (1+\sin^2x (...)))$$

I don't know how interesting or useful that is, just something to think about I guess haha.

anonymous · Answer

$$1+\sin^2x(1+\sin^2x(....) = y \1+(\sin^2x)(y) =y \ y-ysin^2x= 1 \ycos^2x =1\ y= \sec^2x$$

just for the heck of it xD

kainui · Answer

Haha I just finished at the same time as you posted that myself. Fun stuff, actually you know the trig identities for sin^2x, cos^2x, and sinxcosx in terms of a single a trig function? I found a fairly intuitive way of deriving them if you're interested since I hate memorizing them but now I can fairly easily just come up with them in my head lol.

anonymous · Answer

m gonna go now :P.. have fun with math xD

anonymous · Answer

here is something interesting

do u know what is

1+2+3+4..... upto infinity?

anonymous · Answer

The answer is -1/12 :D.. true story!

kainui · Answer

Yeah, but we all know geometric series diverges outside of -1 and 1. ;P

kainui · Answer

S=1+x+x^2+x^3+...
xS=x+x^2+x^3+...
1+xS=1+x+x^2+x^3+...=S
1+xS=S
1=S(1-x)
S=1/(1-x)

=P