Question

Calculus help please?

Evaluate this trigonometric integral:
integral (sin x/ (1-sin^2 x) ) dx

Answer 1

\(\Large\bf\sf 1-\sin^2x\quad=\quad \cos^2x\)
yes?

Answer 2

\[\Large\bf\sf \int\limits \frac{\sin x}{\cos^2x}\;dx\quad=\quad \int\limits \frac{1}{\color{royalblue}{\cos}^2\color{royalblue}{x}}\left(\color{orangered}{\sin x\;dx}\right)\]We want to make the substitution:
\[\Large\bf\sf \color{royalblue}{u=\cos x}\]Understand how to proceed from there?

Answer 3

Ok I can see what you did .... do 1/ cos^2x becomes cos x? or does (sin x dx) become cos x?

Answer 4

@zepdrix ?

Answer 5

what?
I'm not sure what you're asking..

Answer 6

\[\Large\bf\sf \int\limits\limits \frac{1}{\color{royalblue}{\cos}^2\color{royalblue}{x}}\left(\color{orangered}{\sin x\;dx}\right)\quad=\quad \int\limits\limits \frac{1}{\color{royalblue}{u}^2}\left(\color{orangered}{\sin x\;dx}\right)\]The substitution changes the problem to this, understand?
We still need to deal with the sin x dx

Answer 7

ok I'm sorry now I see

Answer 8

well sin x dx = - cos x + C right?

Answer 9

No we don't integrate the sin x dx.
We can't integrate something that involves `both` u and x.

Answer 10

We need to change `everything with x in it` to `something with u in it`.
We replaced the cosx with u.

Now we need to replace sinx dx with something involving u, BEFORE we can integrate.

Answer 11

To find a replacement, take the derivative of your substitution,\[\Large\bf\sf \color{royalblue}{u=\cos x},\qquad\qquad\qquad \frac{du}{dx}=?\]

Answer 12

- sin x?

Answer 13

or 0 my bad... I think

Answer 14

Ok good, your first guess looks right. We'll "multiply" the dx to the other side, giving us:\[\Large\bf\sf du=-\color{orangered}{\sin x \;dx}\]
See how we `almost` have something to replace with sin x dx in our integral?

Answer 15

ok so my first guess was right :)

Answer 16

ok

Answer 17

We want to "solve for" sin x dx.
See how the negative sign is in the way?
What can we do about that? :o

Answer 18

tranfer it to du?

Answer 19

Multiply the negative to the other side?
Ok good.\[\Large\bf\sf \color{orangered}{-du=\sin x \;dx}\]Ok ok ok.
See what we can plug in for our sinx dx now?

Answer 20

hmm I think thats - cos ^-1 ?

Answer 21

Ok looks good!
But `do not` use -1 for powers of trig functions.
We reserve that for the inverse function.\[\Large\bf\sf \cos^{-1}x\quad\ne\quad \frac{1}{\cos x}\]

You can write it like this if you want,\[\Large\bf\sf (\cos x)^{-1}\]
I think your negative goes away when you integrate, yes?
You picked up a -1 from the sinxdx substitution.
Then you lose it when you integrate, gaining a factor of -1 from the u.

Answer 22

ok so 1/cos x is my final answer right?

Answer 23

Mmm ya that sounds right! good job.

Answer 24

thank you soo much!!