Question

\[Evaluate \int\limits \frac{\sin3x}{\cos ^{5}3x}\]

Answer 1

u = cos 3x

Answer 2

see what happens

Answer 3

\[-\int\limits \frac{du}{u ^{5}}\]

Answer 4

is it correct?

Answer 5

yes

Answer 6

missing a 3

Answer 7

what's next?

Answer 8

yup typo error.

Answer 9

missing a 3???

Answer 10

\[\large \int\limits u^n du=\frac{u^{n+1}}{n+1}+C\]

Answer 11

yes u missed 3 because du =-3 sin 3x dx

Answer 12

after letting u=cos 3x u must have
\[\int\limits -\frac{1}{3 u^5} du\]

Answer 13

okay thanks. i get it. :)

Answer 14

yw :D

Answer 15

@mukushla what is integ of \[\int\limits -\frac{1}{3u ^{5}}\]

Answer 16

see my hint up there ; here is n=-5 try to work it out
and im here if u have any question

Answer 17

thank you so much. :)

Answer 18

yw my friend