Question

Find the Integral
Integral of (pi/2 , 3pi/2) Cosx/Sin^2x dx

Answer 1

Hi :)

did you try u = sin x
?

du =...?

Answer 2

let me try that yes, I totally forgot I had that option. Um, u= sin^2x ? Or should I omit the ^2? and the du= cosxdx?

Answer 3

yessss

u = sin x

du = cos x dx is correct


now change the limits too
when x =pi/2
u=...?

when x= 3pi/2
u=...?

Answer 4

x= pi/2 -> u= sin(pi/2) ? and then the same with 3pi/2?

Answer 5

yes,

whats u = sin(pi/2) = ...?
whats u = sin(3pi/2) = ...?

Answer 6

1 and -1

Answer 7

yes, so your new limits are from 1 to -1

Answer 8

whats your new integral ?

Answer 9

integral of (-1, 1) u^2 / du ?

Answer 10

u = sin x

and sin x is in denominator , so
\(\Large \int \limits_{-1}^1\dfrac{1}{u^2}du\)
right ??

Answer 11

yes, then substitute with the u values, and then substitute with the integral values ?

Answer 12

integrate 1/u^2
can you ?

Answer 13

1/ sinx^2 so then it would be sin(-1)^2

Answer 14

there is no sin x now...

Answer 15

your new integral is just in u

Answer 16

how is that? u= -1?

Answer 17

here n = -2