Question

Evaluate this integral

Answer 1

\[\int\limits_{?}^{?} \frac{ secx }{ sinx +cosx } dx\]

Answer 2

So far I did...\[secx = \frac{ 1 }{ cosx }\]

\[\int\limits_{?}^{?} \frac{ 1 }{ sinxcosx + \cos ^{2}x }\]

Answer 3

what are your limits?

Answer 4

there are none its not a definite integral....

Answer 5

I would use the method of substitution,
We can replace
\[\frac{ \sec(x) }{ \cos(x)+\sin(x) } = \frac{ \sec ^{2}(x) }{ 1+\tan(x) }\]
Then using substitution of u=tan(x) it can be simplified quite dramatically

I won't tell you the answer but I'll help you get there :)

Answer 6

Well \[\sec ^{2}x = 1 + \tan ^{2}x\]

Can we write

\[\int\limits_{?}^{?} \frac{ 1 + \tan ^{2}x }{1 + tanx }dx\]
?@sarahusher

Answer 7

you could, but that would be making it more complicated than needs be
what is the differential of tan(x)?

Answer 8

Is it sec^2x ?

Answer 9

yep :)
so if you use
u=tan(x)
you know\[\frac{ du }{ dx }=\sec ^{2}x\]
if you substitute all of this is, what does it come to?

Answer 10

Its \[\int\limits_{?}^{?}\frac{ 1 }{ u } du\]
\[lnu + C\]
\[\ln (1 + \tan(x)) + C\]

right?

Answer 11

okay you should get\[\int\limits_{}^{}\frac{ 1 }{ 1+u }du\]

Answer 12

Why can't \[u = 1 + tanx\]
instead of \[ u = tanx\]

Answer 13

I thought that the formula for the arctan was \[\int\limits_{?}^{?}\frac{ 1 }{ 1 + u ^{2} }\]

Answer 14

Okay, well I think your method works too (i'm not saying it doesnt!!)
just with the route i was taking you down
i was doing \[\int\limits_{?}^{?}\frac{ 1 }{ 1+u }du\]
but if you integrate it via your route then we can compare answers which is great!

Answer 15

Can you complete your route for me please @sarahusher I want to know where you were taking this ?

Answer 16

Yes of course I can:
let tan(x)=u
So \[\int\limits_{?}^{?}\frac{ 1 }{ 1+u }du\]
If you integrate you get
ln(1+u) + C
which you can sub back in and get
ln(1+tan(x))+c

Answer 17

Oh sweet.. Same thing I did thanks @sarahusher haha.... I got a lot more ill post if I need help...

Answer 18

Great, I'm glad!!
Yes please do, i'll help as much as i can, and if i can;t i'll find someone who can! :)