Question

Evaluating an integral.

Answer 1

\[I=\int\limits_{0}^{\pi/2}(\frac{ 1 }{ 1+\tan^{6,000}\theta})d(\theta)\]
and
\[J=\int\limits_{0}^{\pi/2}(\frac{ 1 }{ 1+\cot^{6,000} \theta})d(\theta)\]

1) Evaluate limits
2)Show J = I
2) what is J+I?

Answer 2

f(x)=f(a-x) when limits are from 0 toa..u know this?

Answer 3

emm no, I didn't know o.0

Answer 4

\[\int\limits_{0}^{a} f(x) = \int\limits_{0}^{a}f(a-x)\]

Answer 5

using this u can easily prove 2)

Answer 6

hmm haha you'd have to show me.
is f(x) = I ?

Answer 7

here a=pi/2 and we know tan(pi/2-θ)=cot θ..now u get it?

Answer 8

no not really ...how do we know the value for theta?

Answer 9

we needn't know it..just apply the above two properties and u will see J=I

Answer 10

\[\int\limits_{0}^{\pi/2} 1/(1+\tan ^{6000}((\pi/2)-θ))\] =??

Answer 11

hint: use tan to cot conversion that i have mentioned above

Answer 12

...there are you evaluating at the end points?

Answer 13

tan(pi/2-θ)=cot θ
how do we know this?

Answer 14

its a formula (TRIG. CONVERSIONS)

Answer 15

ok?got it?

Answer 16

how did you plug pi/2 into that integral?

Answer 17

what does it mean when they "evaluate limits"?

Answer 18

I think to solve them.

Answer 19

solve what limits?

Answer 20

I mean do they mean integrals?

Answer 21

She was doing limit definition for integration earlier...
maybe that's what they're going on about.. I dunno

Answer 22

part 2 and 3 make sense
but I have no clue what part 1 means

Answer 23

oh sorry!! typo. I meant integrals xD

Answer 24

I think we can easily find part 1 by first looking at part 2 and 3

Answer 25

have you did part 2 yet?

Answer 26

@priyab2 said something about using a co-function identity and doing a substitution for part 2 which does work very awesomely

Answer 27

ya fun stuff :)

Answer 28

\[\int\limits_0^\frac{\pi}{2} \frac{1}{1+(\cot(\theta))^{6000} } d \theta \\ \cot(\theta)=\tan(\frac{\pi}{2}-\theta) \text{ is a co-function identity } \\ \int\limits_0^\frac{\pi}{2} \frac{1}{1+(\tan(\frac{\pi}{2}-\theta))^{6000}} d \theta \\ \text{ use a substitution on the } \frac{\pi}{2}-\theta \text{ part }\]

Answer 29

Yeah, I got part 2:)

Answer 30

so we have I=J then
if you can find I+J then you can easily find I and J separately which would be part 1

Answer 31

big hint for part 3 just add the integrals as they were

Answer 32

find a common denominator and combine fractions

Answer 33

you will see something really cool happen

Answer 34

add them as they were before the u sub?

Answer 35

their original forms?

Answer 36

yep

Answer 37

hmm would it be easier to do that after evaluating?

Answer 38

no you don't want to evaluate the integrals

Answer 39

force a common denominator
\[\frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{db}\]

Answer 40

\[\frac{1}{1+\tan^{6000}(\theta)}+\frac{1}{1+\cot^{6000}(\theta)}=?\]

Answer 41

\[\frac{ (1+cot^{6000}(theta))+(1+tan^{6000}(theta)) }{ (1+\tan^{6000}\theta)(1+\cot^{6000}(\theta) }\]..?

Answer 42

okay multiply the bottom out...

Answer 43

you do know that tan(x)*cot(x)=1 on tan(x)*cot(x)'s domain right?

Answer 44

\[\tan(x)\cot(x)=1 \\ \text{ so } (\tan(x)\cot(x))^n=1^n \\ \text{ gives } \tan^n(x) \cot^n(x)=1\]

Answer 45

k just a sec, i'm writing it all out haha

Answer 46

man these ^6000 are messing me up. I am quite disoriented.

Answer 47

I know it's like a normal ^n haha but still D:

Answer 48

(1+tan^n(theta))*(1+cot^n(theta))
I would just foil it out, yeah?

Answer 49

\[(1+\tan^n(x))(1+\cot^n(x)) \\ 1(1+\cot^n(x))+\tan^n(x)(1+\cot^n(x)) \\ =...\]
distribute

Answer 50

yeah you can use that thing that is called foil

Answer 51

(tan^n)(cot^n) = 1?

Answer 52

yes

Answer 53

well tan^n(x)*cot^n(x)=1 yes

Answer 54

so we'd just end up with
1+tan^(6000)(theta)+1+cot^(6000)(theta)

Answer 55

and what do we have on the numerator ?

Answer 56

(1+cot^(6000)(theta))*(1+tan^(6000)(theta))

Answer 57

:(

Answer 58

should be + not *

Answer 59

oh I guess I should simplify that one too

Answer 60

oops, that's what I meant :}

Answer 61

so if you have the same thing on top and bottom...

Answer 62

0.0 it..does it equal 1?

Answer 63

\[I+J=\int\limits_0^\frac{\pi}{2} 1 d \theta \]

Answer 64

mind..blown. thoroughly.

Answer 65

you got it from here right?

Answer 66

now I evaluate at end points

Answer 67

you integrate 1
and yes plug in the endpoints

Answer 68

which just = pi/2

Answer 69

\[I+J=\frac{\pi}{2} \\ I=J \\ \text{ so } I+I=\frac{\pi}{2} \implies I=?\]

Answer 70

I = (pi/2)-I ?

Answer 71

I+I=2I :p

Answer 72

then you solve for I

Answer 73

come on
I know your algebra solving skills are better than that :p

Answer 74

you solved equations like x+x=5 or something like that right?

Answer 75

x+x=5
2x=5
and so x=5/2

similarly you can solve I+I=pi/2

Answer 76

hahah #braindead.
pi/4 yeah?

Answer 77

yeah
and since I=J then J is also pi/4

Answer 78

so you have found what both definite integrals equal now

Answer 79

and that solves part 1? :o

Answer 80

yes it is weird they put that as part 1 in my opinion

Answer 81

wow that was quite glorious.

Answer 82

probably for the students to spend forever trying to evaluate it in it's original form xD which I tried haha

Answer 83

unless that was suppose to be the main question and they were using the other questions to lead you there

Answer 84

you can easily show that one co-function identity you weren't convinced was an identity if you want me to show you