Question

JamesJ: would i be able to do the same with this problem??? integral 0-pi/4 sec^2 t dt

Answer 1

I like how you're thinking, rmalik2. :)

Answer 2

what you mean..?

Answer 3

INT sec^t dt = tan t + c
between limits 0 and pi/4 forget c
and calculate tan pi/4 - tan 0

Answer 4

I assume you're asking JamesJ if you can do this:\[\int x^n dx=\frac{x^{n+1}}{n+1}\]with this\[\int_{0}^{\pi}4\cdot\sec^2(t) dt.\]

Answer 5

So the 'trick' here is knowing how to integrate sec^2(x)

Answer 6

yea i know the identity of sec^2t = 1-tan^2t

Answer 7

which is a bit subtle. What is the derivative of f(x) = tan x ?

Answer 8

yes lol

Answer 9

btw, your identity isn't quite right. 1 + tan^2 x = sec^2 x

Answer 10

o sorry 1 + tan^2 x = sex ^2x

Answer 11

Now: What is the derivative of f(x) = tan x ?

Answer 12

calculate using the quotient rule: tan x = sin x / cos x

Answer 13

yes....

Answer 14

f(x)= -log cosx?

Answer 15

no, you integrated. What is the derivative of tan x = sin x / cos x ?

Answer 16

Use the quotient rule

Answer 17

im not sure

Answer 18

Write u = sin x and v = cos x

Then what is the derivative of u/v ?

Answer 19

Do you know the quotient rule?

Answer 20

no i dont

Answer 21

Do you know the product rule?

Answer 22

the product rule for differentiating, that is.

Answer 23

i think so would it be the chain rule??

Answer 24

No, that's another thing.

Answer 25

The product rule says if f is function that can be written as
f(x) = u(x).v(x)
for some other functions u and v. Then if u and v can be differentiated, then the derivative of f is given by

f' = u'.v + u.v'

Answer 26

o well then im not sure

Answer 27

The quotient rule says if we can write
f = u/v

then the derivative of f is

\[f' = \frac{v.u' - u.v'}{v^2}\]

Answer 28

Do these look familiar now?

Answer 29

oooooooo yea now i get that

Answer 30

Ok. Now write tan x = sin x / cos x. What is its derivative?

Answer 31

but im not sure

Answer 32

Let u = sin x and v = cos x. Now apply the quotient rule that I just wrote down.

Answer 33

I know this feels like a long way from your problem, but trust me, we're on the right road and once we turn the corner it will be clear.

Answer 34

Got it yet?

Answer 35

If u = sin x and v = cos x
What are u' and v' ?

Answer 36

u' is -cosx v' sinx

Answer 37

Right. So now apply the quotient rule

Answer 38

to where?

Answer 39

tan x = sin x / cos x = u/v

And if f = u/v, then

\[f' = \frac{v.u' - u.v'}{v^2}\]

Answer 40

I.e., calculate the derivative of tan x.

Answer 41

o so it becomes sinx * -cosx- cos x *sin x/ sin^2x

Answer 42

No, you've got everything backwards. Look again at what u and v are.

Answer 43

well i did say u'=-cosx and v'=sinx...so (cos x)(-cos x)-(sinx)(sinx)/cos^2x?

Answer 44

Oh, there' the problem. If u = sin x, u' = cos x
If v = cos x, v' = -sin x

Answer 45

i thought its -cos x

Answer 46

Hence for f(x) = tan x

\[f' = \frac{\cos x . \cos x - \sin x.(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x}\]

Answer 47

i.e., f'(x) = sec^2 x.

Answer 48

Hence applying this result to your original problem and using the fundamental theorem of calculus, the indefinite integral of sec^2 x is tan x.

Answer 49

yes but why go through all this when i know that it would be tan pi/4 -tan(0)=1?

Answer 50

The question is HOW do you know that the integral of sec^2 x is tan x?

Answer 51

...because the problems won't always be this easy and you need to see how this works so when a sophisticated variation comes along, you can figure it out.

Answer 52

well sinx+ cosx=1 and when i divid sinx i would get 1+tanx=1/secx right?

Answer 53

What?!

Answer 54

sin^2 x + cos^2 x = 1

Answer 55

yes i meant that i just didnt put the ^2 on

Answer 56

and you divide by cos^2 x to get

tan^2 x + 1 = sec^2 x

Answer 57

But this last identity has nothing to do with your problem.

Answer 58

ok