Question

What is the antiderivative of sinx/cos^2x?

Answer 1

clarification: antiderivative of sinx/cos^2(x)

Answer 2

@zepdrix

Answer 3

i don't understand the u substitution method properly

Answer 4

oops, i meant try \(u=\cos(x), du=-\sin(x)dx\)

Answer 5

it is the backwards chain rule

Answer 6

so you get -du/u^2?

Answer 7

yes, and lets finish it, check that it is right, and see why the method works

Answer 8

did you get \[-\int \frac{du}{u^2}=\frac{1}{u}\] yet?

Answer 9

It might be nice to just rewrite it as \[\Large \int\limits \frac{\sin x}{\cos^2x}dx = \int\limits \frac{\sin x}{\cos x}\frac{1}{\cos x}dx=\int\limits \tan x \sec x \ dx\]

Answer 10

ok yea i got -1/u

Answer 11

@Kainui more than one way to skin a cat for sure

Answer 12

hmm you are off by a minus sign i think

Answer 13

oh yeah i typed that by accident :/
sorry

Answer 14

so then 1/u=1/cosx

Answer 15

so it will be \(\frac{1}{\cos(x)}\) or as @Kainui pointed out \(\sec(x)\)

Answer 16

now why is that right? if you take the derivative of \(\frac{1}{\cos(x)}\) you need the chain rule
you get
\[-\frac{1}{\cos^2(x)}\times -\sin(x)\]

Answer 17

so the derivative of the answer is equal to the original function

Answer 18

the chain rule for taking derivatives gives you the "derivative of the outside evaluated at the inside, times the derivative of the inside"
u - sub works that routine backwards

Answer 19

so to check my answers I can just take the derivative of the answer?

Answer 20

\[\int\limits_{}^{}f(x) dx =F(x)+C \text{ if } (F(x)+C)'=f(x)\]

Answer 21

yes that is what you are looking for, the function whose derivative is between the \(\int\) and the \(dx\)

Answer 22

ok thank you!

Answer 23

yw