Question

Definite integrals

Answer 1

a function of which a given function is the derivative, i.e., which yields that function when differentiated, and which may express the area under the curve of a graph of the function.

Answer 2

Have you considered a substitution? \(u = \cos(x)\), maybe?

Answer 3

@willet : Yes, @tkhunny is correct.

Answer 4

@tkhunny \[-\int\limits_{-1}^{1} \frac{ 1 }{ 16+u^2 }\]

@tkhunny thats where im at, I think i did everything correctly

Answer 5

Okay, now about about another substitution?

\(\tan^{2}(v) + 1 = \sec^{2}(v)\)

Answer 6

Where did you get that from?

Answer 7

It's an identity you should know. If you need to brush up on your trigonometry, this would be a good time to do it. You will need it.

This one is easy to create on your own. Start with the Pythagorean Identity in sine and cosine and divide by cos^2(x).

Answer 8

If you let your initial substitution be,\[\Large\rm 4u=\cos x\]I think it will work out a smidge easier.

You get something that looks a lot closer to the arctangent integral thing.

Answer 9

im lost

Answer 10

Right. There was no need to do two substitutions. It may have helped you see it all the way through, I suppose.

You could do \(4\cdot\tan(u) = \cos(x)\) and go straight to it!

Answer 11

I have to get it in the form of \[\int\limits \frac{ du }{ a^2+u^2 }\] right

Answer 12

@nincompoop can you help? im lost

Answer 13

you already have it in that form. a=4 from your earlier post. do you remember how to do trig sub? i can tell you right now that will be an arctan

Answer 14

and you forgot to put your du on top earlier. what calc are you in, 1 or 2?

Answer 15

calc 1

Answer 16

ok then your professor probly taught you a formula and said memorize it. I suggest you learn how to derive it if you are going higher than calc 1. anyway i dont have it memorized, but i can show you how to get the answer without the formula. or look in your book on the first page and it tells you the formula

Answer 17

Once I have it in that form, I put it in 1/a(arctan)u/a + C

Answer 18

k so your a=4(let me know if you dont know why), what else do you know?

Answer 19

I'm not sure how the technical way of saying it, but a = 4 because a^2 = 16

Answer 20

and I have u = cosx

Answer 21

yup, what do you get when you plug and chug in the formula?

Answer 22

its gonna be 1/4(arctan(cosx/4)) and its not + C. Can you tell me why?

Answer 23

why not?

Answer 24

because this is a definite integral. we are given limits from 0 to pi so we have to evaluate it. and I forgot a negative btw.

Answer 25

Oh ok

Answer 26

alright im going to bed, any last questions?

Answer 27

-1/4( arctan(cos(pi)/4) - arctan(cos(o)))

Answer 28

Wouldn't that be cos(0)/4? at the end?