Question

integration

Answer 1

find
\[\Huge \int\limits \frac{\cos^2x\;dx}{\cos^2x+9\sin^2x}\]

Answer 2

i tried dividing with \[\large \cos^4 x\]

Answer 3

bt still nt reching the crct place

Answer 4

*reaching

Answer 5

Multiply and divide by \(\cos^2x\) then, substitute \(t=\tan x\).\[\frac{1}{(1+t^2)(1+9t^2)} dt\]

Answer 6

hw u gt that?

Answer 7

\[\frac1{(1+t^2)(1+9t^2)} = \frac{At +B}{1+t^2} + \frac{Ct + D}{1+9t^2}\]
I hope this is right.

Answer 8

I already told you lol.

Answer 9

where is the multiplied\[\cos^4 x\]?

Answer 10

\[\frac{1}{(1+t^2)(1+9t^2)}= \frac{1}{-8} \left(\frac{1}{1+t^2} - \frac{9}{1+9t^2}\right)\]

Answer 11

\[\Large\frac{\frac{\cos^2x}{\cos^2x}}{\frac{\cos^2x + 9 \sin^2x}{\cos^2x}}\]

Answer 12

ooh !!

Answer 13

I did it with \(\cos^2x\) not \(\cos^4x\)

Answer 14

i was confused by ur statement multiply and divide ..i was looking where is multiplication lols

Answer 15

cant we get to it by dividing with \(\huge \cos^4 x?\)

Answer 16

you can but i don't want to

Answer 17

hw to integrate \[\huge \frac{1}{1+9t^2}\]

Answer 18

it's 9..the numerator

Answer 19

:) i took 9 outside :P

Answer 20

oh well then arctan yup :)

Answer 21

i didnt get hw it is arc tan

Answer 22

helppp

Answer 23

@dpaInc

Answer 24

huh?

Answer 25

hehe...tell me hw it is arc tan

Answer 26

i was referring to \[\huge \frac{1}{1+9t^2}\].
it's in the form 1/(a^2 + u^2)

Answer 27

it's one of the basic inverse trig thingies

\(\LARGE \int \frac{1}{a^2 + u^2} = \frac{1}{a} \tan^{-1} ({u}{a})\)

Answer 28

u is 3t?

Answer 29

\(\LARGE \frac{1}{a} \tan^{-1} (\frac{u}{a})\)

sorry latex problem

Answer 30

there you go...LGBA^

Answer 31

and yes u=3t @AravindG

Answer 32

hm thanks

Answer 33

u welcome