Question

Integrate ∫(2 cos⁡x)/(1+sin^2 x)

Answer 1

is it \(\huge \int \frac{2cos x}{1+sin^2 x}\) ?

Answer 2

if yes, u can put t=sin x

Answer 3

Yeah...i get it! Thank you v much :)

Answer 4

welcome ^_^

Answer 5

any more doubts / questions ?

Answer 6

Umm..yes...integration is driving me crazy T.T
What about this question...
Integrate 1/ 1+sin^2 X
use the same approach?

Answer 7

hmm... no
u divide numerator and denominator by cos^2 x
and then put t=tan x
if u see my tutorial, uget a hint on this..

Answer 8

sec^x / (sec^2 x+1)
t= tan x
sec^2 x = 1+t^2
dt=.... ?

Answer 9

Srry, i still cant get it...What i did is like this:
1/(1+sin^2 x)= (1/(cos^2 x))/((1+sin^2 x)/(cos^2 x))
= (sec^2 x)/(sec^2 x+tan^2 x)

Answer 10

oh yes, my mistake
now you can put t=tan x
sec^2 x + tan^2 x= 1+tan^2x+tan^2x = 1+2t^2
and dt=.. ?

Answer 11

So it bcome ∫〖1/(1+2t^2) dt ,am i right?
Then i struck....

Answer 12

right

Answer 13

standard integral
\(\int 1/(1+x^2)dx=tan^{-1}x\)

Answer 14

here first write 2t^2 as
\(\sqrt{2t}\)

Answer 15

i mean \(\sqrt{2t}^2\)

Answer 16

so u finally get \(tan^{-1}(\sqrt2 t)/\sqrt2+c\)

Answer 17

This is what i did:
∫1/(1+(√2 t)^2) dt
Let u=√2t
dt=1/√2 t
(1/√2) ∫1/(1+u^2) du
(1/√2 ) tan^(-1)⁡u
(1/√2) tan^(-1)⁡〖√2 t〗
and finally, i get this (1/√2 ) tan^(-1)⁡〖√2 tan⁡x 〗
Any mistake here?

Answer 18

thats correct

Answer 19

Alright.Thx a lot. U really help me a lot...:)

Answer 20

welcome ^_^