Question

PLEASE HELP! URGENT! MEDAL! @solomonzelman @Hero @ganeshie8

Answer 1

so far I've simplified it to tan(t)sec(t)

Answer 2

@zepdrix @SolomonZelman @mathmale

Answer 3

Using the tan(t)sec(t)dt simplification, I would try integration by parts where dv = sec(t)dt and u = tan(t).

Answer 4

??? I'm bad at u substitution. Can you show me?

Answer 5

\(\color{#000000 }{ \displaystyle \int{\frac{\sin(x)}{1-\sin^2{x}}dx} }\)

\(\color{#000000 }{ \displaystyle \int{\frac{\sin(x)}{\cos^2{x}}dx} }\)

u=cos(x)

Answer 6

so...you'd get sin(x) / u^2

Answer 7

Don't forget that du = -sin(x).

Answer 8

?

Answer 9

\(\color{#000000 }{ \displaystyle \int \frac{f'(x)}{f(x)}dx }\)

If you are setting
\(\color{#000000 }{ \displaystyle u=f(x) }\)

Then you have a to make a "chain rule" for "u"
(If you think about differentiation, this must be logical)

\(\color{#000000 }{ \displaystyle \frac{du}{dx}=f'(x) }\) (derivative of f(x) is f'(x), ok?)

\(\color{#000000 }{ \displaystyle du=f'(x)dx }\) (yes, we could manipulate differentials like this)

And notice that:
\(\color{#000000 }{ \displaystyle \int \frac{1}{\color{blue}{f(x)}}\color{red}{f'(x) dx} \quad \Longrightarrow \quad \int \frac{1}{\color{blue}{u}}\color{red}{du} }\)

Answer 10

Also, anytime you have

\(\color{#000000 }{ \displaystyle \int f'(x)\cdot G(f(x))dx }\)

If you are setting
\(\color{#000000 }{ \displaystyle u=f(x) }\)

\(\color{#000000 }{ \displaystyle \frac{du}{dx}=f'(x) }\) (derivative of f(x) is f'(x))

\(\color{#000000 }{ \displaystyle du=f'(x)dx }\)

And notice that:
\(\color{#000000 }{ \displaystyle \int \color{red}{f'(x)}\cdot G(\color{blue}{f(x)})\color{red}{dx} \quad \Longrightarrow \quad \int G(\color{blue}{u})\color{red}{du} }\)

Answer 11

ok, so if I set u = sin(x), then for du, you'd get cos(x)dx, right?

Answer 12

u=cos(x)

Answer 13

oh, ok.

Answer 14

\(\color{#000000 }{ \displaystyle \int{\frac{\sin(x)}{1-\sin^2{x}}dx} }\)

\(\color{#000000 }{ \displaystyle \int{\frac{\sin(x)}{\left(\cos{x}\right)^2}dx} }\)

u=cos(x)

Answer 15

Similar to G( f(x) ) form that I showed... isn't it?

Answer 16

so du = -sin(x)dx

Answer 17

**so far I've simplified it to tan(t)sec(t)**

i'd argue that that's one you should probably carry around in your head

Answer 18

"recognizing derivatives" :)

Answer 19

is that right @SolomonZelman

Answer 20

yes,

u=cos(x)

du= -sin(x) dx
and since you don't have the negative in the integral you can write,
-du=sin(x) dx

Answer 21

ok, now what?

Answer 22

Subtitute these two things into the integral

u=cos(x)
-du=sin(x) dx

Answer 23

\(\color{#000000 }{ \displaystyle \int{\frac{\color{red}{\sin(x)dx}}{\left(\color{blue}{\cos{x}}\right)^2}} }\)

Answer 24

but there is no -du in the integral...?

Answer 25

yes, you are substituting the "-du"

Answer 26

there is no "u" in the integral either.
you are substituting it.

Answer 27

Ah!, so you get -du/u^2

Answer 28

Fabulous !

Answer 29

ok, so what's next

Answer 30

\(\color{#000000 }{ \displaystyle \int{\frac{\color{red}{-du}}{\left(\color{blue}{u}\right)^2}} }\)

Answer 31

re-write with u^{?}
and integral via the power rule

Answer 32

\(\color{#000000 }{ \displaystyle \int (1/u^2) du }\)

\(\color{#000000 }{ \displaystyle \int u^{-2} du }\)

Answer 33

-u^-1

Answer 34

Awesome, but with +C,
and now you need to substitute the x back into the equation

Answer 35

you had
u=cos(x),

so just go ahead and sub

Answer 36

what about the fact that du is negative? does that change it to u^-1?

Answer 37

du is not negative.

Answer 38

but I know what you mean

Answer 39

this is just a matter of rearranging.

we did: u=cos(x)

du/dx =-sin(x)
du=-sin(x) dx
-du = sin(x) dx

we were just rearranging this derivative so that we get the substitution in terms of u, for everything that we have in terms of x. (if you will)

Answer 40

so, basically we treat du sorta like 0, and the - disappears?

Answer 41

well, when you integrate, you remove the du and the integral sign, but put the +C in indefinite integral.

Answer 42

so yes, you basically treat like zero, if you will.... ((but it isn't really zero, ain't it?))

Answer 43

ok, awesome. Thank you so much. Can you help me with one more?

Answer 44

ure

Answer 45

sure

Answer 46

Oh, btw, one more thing

Answer 47

yea?

Answer 48

just clarifying that we did get u^1 => 1/cos(x), right?

Answer 49

yep

Answer 50

i tagged you to my next question

Answer 51

Irishboy said this, but we might have both ignored this.
(the answer we got is sec(x) )

\(\color{#000000 }{ \displaystyle \int{\frac{\sin(x)}{1-\sin^2{x}}dx} }\)

\(\color{#000000 }{ \displaystyle \int{\frac{\sin(x)}{\cos^2{x}}dx} }\)

\(\color{#000000 }{ \displaystyle \int{\frac{\sin x}{\cos{x}}\times \frac{1}{\cos x} dx} }\)

\(\color{#000000 }{ \displaystyle \int{\tan x \sec x dx} }\)

and based on the fact that

\(\color{#000000 }{ \displaystyle \frac{d}{dx} (\sec x) =\tan x \sec x }\)

We get that

\(\color{#000000 }{ \displaystyle \int{\tan x \sec x dx}=\sec x \color{grey}{ +C} }\)