Question

what is the integral of [csc(sinx)]^2 cosx dx, when u=sinx

Answer 1

what did u get after plugging in u= sin x ??
du =.. ?

Answer 2

cosx

Answer 3

du = cos x dx to be precise
so, whats your integral now ?

Answer 4

in terms of u only

Answer 5

[cscx(u)]^2 du..

Answer 6

correct!
can you find that integral ? its standard

Answer 7

do i break up the ^2 for the u and csc x?

Answer 8

do you know the derivative of cot x ?

Answer 9

csc^2 x

Answer 10

its - csc^2 x right ??
\((d/dx)\cot x = -\csc^2x \\ \implies \int \csc^2xdx = -\cot x +c\)
got this ?

Answer 11

yea i got it.

Answer 12

thats it!
final answer is -cot x +c :)

Answer 13

now only thing remain is to plug back in for 'u'

Answer 14

-cot(sinx) +c

Answer 15

oh, typo, so you got
-cot u+c right ??

Answer 16

and u = sin x gives you
-cot (sin x) + c

Answer 17

yes :)

Answer 18

yea thats what i got. Thanks!

Answer 19

welcome ^_^

Answer 20

one question. why didnt we square the sin x. we only did the csc x?? @hartnn

Answer 21

we plugged in u = sin x
so we had
csc^2 u du

where was the need to square the sin x ?

Answer 22

but isnt the square for both csc(sinx) so we distribute the ^2 to both terms right?
so we plugged in u for sin.
so why not csc^2 u^2?

Answer 23

oh, its not PRODUCT of function !
not like (ab)^2 = a^2 b^2

Answer 24

its composite function,
csc (sin x) ------> like f(g(x))

Answer 25

and [f(g(x))]^2 = f^2 (g(x))

Answer 26

ooh. now it makes more sense! thanks again :)

Answer 27

welcome ^_^