Question

how do i find the indefinate integral of -sin(x)/cos^2(x)dx?

Answer 1

\[\Large\bf\sf \int\limits \frac{1}{\cos^2x}\left(-\sin x\;dx\right)\]

If I split it up like that, can you see your u-substitution easier?

Answer 2

i am not sure

Answer 3

what do i do from there?

Answer 4

\[\Large\bf\sf u=\cos x\]

Answer 5

What do you get for `du` ?

Answer 6

-cos(x)?

Answer 7

No silly!
We take our substitution and then take its derivative with respect to x.\[\Large\bf\sf u=\cos x, \qquad\implies\qquad \frac{du}{dx}=-\sin x\]Then simply "multiply" your dx to the other side to solve for du.\[\Large\bf\sf du=-\sin x\;dx\]

Answer 8

\[\Large\bf\sf du=(-\sin x\;dx)\]The du is showing up somewhere in our integral, yes?

Answer 9

yes

Answer 10

\[\Large\bf\sf \int\limits\limits \frac{1}{(\color{royalblue}{\cos x})^2}\left(\color{orangered}{-\sin x\;dx}\right)\]

\[\Large\bf\sf \color{royalblue}{u=\cos x}, \qquad\qquad\qquad \color{orangered}{du=-\sin x\;dx}\]
Do you understand how we'll be plugging in the pieces for u?

Answer 11

yes i s

Answer 12

yes i see it

Answer 13

Ok good.
\[\Large\bf\sf \int\limits\limits\limits \frac{1}{(\color{royalblue}{\cos x})^2}\left(\color{orangered}{-\sin x\;dx}\right)\quad\to\quad \int\limits\limits\limits \frac{1}{(\color{royalblue}{u})^2}\left(\color{orangered}{du}\right)\]

Answer 14

From here we can apply a rule of exponents to write it like this,\[\Large\bf\sf \int\limits u^{-2}\;du\]And then just apply your power rule for integration.

Answer 15

Very nice job at explaining this @zepdrix!
Keep it up :)

Answer 16

color ftw ^^

Answer 17

okay thank you, i think i got it from here, :)

Answer 18

Cool c: