Question

integration sin^2 theta / cos^2 theta * sin^2 theta

Answer 1

\[\int\limits_{}^{} \frac{ \sin^2 \theta }{ \cos^2 \theta \sin^2 \theta }\]

Answer 2

can i cancel out the all the squre?

Answer 3

first simplify sin2(th)

then note that
1/cos^2(th) = 1+tan^2(th)
so
you have
int(1+yan^2(th))d(th)
now its easy

Answer 4

answer is tan(theta) +c(constant)

Answer 5

after the simplify, i got 1/cos^2(th) and i changed this as int sec^2(th)

and now, its tan(th) +c ?

Answer 6

you said, 1/cos^2(th) = 1+tan^2(th) --------> this is also equal to sec^2(th) right?

Answer 7

i got the same answer as you

Answer 8

yes
1/cos^2(th) = 1+tan^2(th)=sec^2(th)

Answer 9

but then i have limit 0 to 2pi ---> so tan(th) is 0?

Answer 10

* int than(th) is 0?

Answer 11

wait a mooment

Answer 12

no the answer is not 0
because function is positive in(0-2pi)

Answer 13

answer of this int (0 - 2pi) is +infiity

Answer 14

hmm its hard.. hahaha

Answer 15

actually this problem is from double integrals in polar coordinates and that i asked the first was just the part of it

Answer 16

is that mean limit 0 to 2pi should be changed?

Answer 17

is it your home work ?
area goes to infinity

Answer 18

because of the vertical retricempote x=pi/2, 3pi/2