Question

Evaluate the indefinite integral of:

1
----
1 + (cosx)^2


I would like only a hint(s) please.

Answer 1

U-sub, integ by parts.... ?

Answer 2

Any ideas Zarkon?

Answer 3

Use a trig identity to make it possible

Answer 4

You mean in the numerator? I would get sin^2(x) + cos^2(x)

Answer 5

But how would that help at all?

Answer 6

There is another identity, should it be revealed to you?

Answer 7

Wow who knew a question could be so popular

Answer 8

\[\int \frac{1}{1+\cos^2 x} dx\]
multiply numerator and denominator by \(\sec^2 x\)
\[\int \frac{\sec^2 x}{1+\sec^2x} dx\]
\[\int \frac{\sec^2 x}{\tan^2x } dx\]
put tan x= t
\[\int \frac{dt}{t^2} \]

Answer 9

Wow Ash, you got some ingenuity there!

Answer 10

Is there any other way?

Answer 11

\[\frac{1}{1+\cos^2(x)}=\frac{\sec^2(x)}{1+\sec^2(x)}=...\]

stole my thunder ;)

Answer 12

though...you hace some mistakes

Answer 13

Wow. I had no idea. Ok I will remember that: multiple by the inverse identity. Check.

Answer 14

@zarkon Where did I make mistakes?:D

Answer 15

\[1+\sec^2(x)=1+1+\tan^2(x)=2+\tan^2(x)=\frac{1}{2}\left(1+\frac{\tan^2(x)}{2}\right)\]

Answer 16

@Zarkon Oh thanks for pointing out
\[\int \frac{ \sec^2 x}{2+\tan^2 x} dx\]

Answer 17

\[1+\sec^2(x)=1+1+\tan^2(x)=2+\tan^2(x)=2\left(1+\frac{\tan^2(x)}{2}\right)\]

Answer 18

had a little typo myself ;)

Answer 19

Where did you get that 2 from?

Answer 20

@QRAwarrior Sorry made a mistake:(
Did you get it?
\[\sec^2 x= 1+ \tan^2 x\]
\[1+\sec^2 x=1+ 1+ \tan^2 x\]

Answer 21

Got it.

Answer 22

Wow are you people mathematicians or what? That was brilliant.

Answer 23

@Zarkon is a real mathematician.:D
Glad that we have him here:D