Question

\[tan^2 \theta +100 = ?\]? in terms of \[sec\theta\]

Answer 1

well i guess thee question is wrong :P

Answer 2

See the Identity is:
\[1 + \tan^2 \theta = \sec^2 \theta\]

Answer 3

so what if its 100 in stead of 1?

Answer 4

Multiply 10 both sides of the formula that I have written..

Answer 5

i see...that would make \[10 + 10\tan^2 \theta = 10\sec^2 \theta\]
so where from here? because I guess I don't know what
\[100 + \tan^2 \theta = \] in terms of sec

Answer 6

nw u have 2 equations for 10sec^2theta,so over solve...

Answer 7

No no go slowly..

See original and formula equation have now become:

\[\tan^2 \theta + 100 = 10\sec^2 \theta\] --------1
\[10\tan^2 \theta + 10 = 10 \sec^2 \theta\] ----2

Now RHS of both the equations are equal so equate LHS..
Getting?

Answer 8

shoot me

Answer 9

i 'm not seeing it....what's wrong with me?

Answer 10

Okay where is the gun??

Answer 11

LOOOOLzzzz!!!!!

Answer 12

Just go with more simple method..

See the formula I have written above..

You know from there the value of sec^2 theta...

Answer 13

you mean substitute one equation into the other and solve for sec^2theta?

Answer 14

See the formula is:

\[1 + \tan^2 \theta = \sec^2 \theta\]

This value of sec^2 theta just put in the question see what you get?

Answer 15

hold on a sec

Answer 16

Take your time..

Answer 17

Have you changed the question??

Answer 18

Well it's the same question....I just don't know what the RHS equals

Answer 19

But there you have written 10sec^2 theta earlier??

Answer 20

10 sec^2theta was my guess

Answer 21

I followed it with a question mark

Answer 22

LOL! Sorry I guess I'm confusing the both of us

Answer 23

So, the original question is:

\[\tan^2 \theta + 100 = ???\]

Is it??

Answer 24

yes....do you want me to write the whole question....this is just a small part of it

Answer 25

definitely need the whole question here

Answer 26

Yes if you want to then write the whole..

Firstly here subtract 100 from both the sides and tell me what you got??

Answer 27

\[\int \frac{x^3}{\sqrt(x^2 +100)} dx\]

Answer 28

LOL no....what does it equal in terms of sec?

Answer 29

you need to remember the constants in your substitutions!
you forgot yours

Answer 30

x=tan theta
dx=sec^2 theta d theta

Answer 31

i was trying to change the denomitor in terms of sec to get rid of the sqrt

Answer 32

yeah, but what about the 100?
we need to take care of that
the sub you want for integrals of the form\[\sqrt{ax^2+b}\]is\[x=\sqrt\frac ba\tan\theta\]look at what happens when you square that and it should make more sense...

Answer 33

hold on turing

Answer 34

ok go ahead...i just don't see how we went to \[x=\sqrt\frac ba\tan\theta\]

Answer 35

because it always works, watch:\[\sqrt{ax^2+b}\]\[x=\sqrt\frac ba\tan\theta\implies x^2=\frac ba\tan^2\theta\]so we get\[\sqrt{a\left(\frac ba\tan^2\theta\right)+b}=\sqrt{b\tan^2\theta+b}=\sqrt b\cdot\sqrt{\tan^2+1}=\sqrt b\sec\theta \]

Answer 36

I can follow this to problem:
\[\sqrt{a\left(\frac ba\tan^2\theta\right)+b}=\sqrt{b\tan^2\theta+b}=\sqrt b\cdot\sqrt{\tan^2+1}=\sqrt b\sec\theta\]


but this is just a given?
\[x^2=\frac ba\tan^2\theta\]

Answer 37

similarly we could write this as\[\sqrt{a^2x^2+b^2}\]\[x=\frac ba\tan\theta\implies x^2=\frac {b^2}{a^2}\tan^2\theta\]so we get\[\sqrt{a\left(\frac {b^2}{a^2}\tan^2\theta\right)+b^2}=\sqrt{b^2\tan^2\theta+b^2}=b\cdot\sqrt{\tan^2+1}=b\sec\theta \]I guess it looks nicer that way

Answer 38

yes, it is a technique that always works; we remember this substitution as the method that will always work, as we have just seen

Answer 39

ok...I'll plug in numbers of a and b and see what happens

Answer 40

no way! 100 sec\theta?

Answer 41

like I said, works every time :)

Answer 42

wait, b for you should be 10

Answer 43

\[tan^2 \theta+100= 10sec^2\theta\]

Answer 44

no that statement is not true...

Answer 45

what we did was say that\[\sqrt{(10\tan\theta)^2+100}=10\sec\theta\]and we did that by making a specific substitution

Answer 46

in your case\[\sqrt{x^2+100}=\sqrt{x^2+10^2}\]we have\(a=1\) and \(b=10\)

Answer 47

\[10tan^2 \theta+100= 10sec\theta\]

Answer 48

still not quite

Answer 49

can you prove that statement above?

Answer 50

make a clear distinction between\[(10\tan\theta)^2=100\tan^2\theta\]and\[10\tan^2\theta\]

Answer 51

\[\sqrt{(10\tan\theta)^2+100}=10\sec\theta\]
\[{(10\tan\theta)^2+100}=100\sec^2\theta\]

Answer 52

we're going in circles
\[{100\tan^2\theta+100}=100\sec^2\theta\]
\[\tan^2\theta +1=sec^2\theta\]

Answer 53

this is not circles... and that is not the right simplification

Answer 54

ok back to the question
\[\int \frac{x^3}{\sqrt(x^2 +100)} dx\]
there must some other way to solve it

Answer 55

really?

Answer 56

we want an expression we can integrate, and to do that we need to get rid of the square root
this trig identity is how we do it, but we need to make the right sub to make it work

Answer 57

Ok

Answer 58

\[(10\tan\theta)^2+100=100\tan^2\theta+100=100(\tan^2\theta+1)=100\sec^2\theta\]

Answer 59

now you have all that under the square root, so that is\[\sqrt{(10\tan\theta)^2+100}=\sqrt{100\tan^2\theta+100}=\sqrt{100}\cdot\sqrt{\tan^2\theta+1}=\sqrt{100\sec^2\theta}\]\[=10\sec\theta\]so hopefully you can see we have gotten somewhere

Answer 60

so you have\[{x^3dx\over\sqrt{x^2+100}}\]we do the sub\[x=10\tan\theta\implies dx=10\sec^2\theta d\theta\]and we get\[{(10\tan\theta)^3{(\cancel{10}\sec^{\cancel2}\theta)}d\theta\over\cancel{10\sec\theta}}=10^3\tan^3\theta\sec\theta d\theta\]

Answer 61

OH MY GOD!!!!!

Answer 62

what, did it click?

Answer 63

Respect

Answer 64

yeah

Answer 65

I'm happy
respect :)

Answer 66

i was trying to change the radical in a different unreasonable way

Answer 67

ya know, I think this integral can be done without a trig sub, but if they really want you to do it that way...

Answer 68

I don't think they need a trig substitution....

but I was trying to change \[\sqrt(tan^2\theta +100)\]
to some form of \[\sqrt(tan\theta)\]....ok i guess life goes on....

Answer 69

lol

Answer 70

but \[x=10\tan\theta\]
instead of tan^2 theta

Answer 71

actually my little trick didn't pan out, but there still may be a way to do this without a trig sub

so can you handle the integral
?

Answer 72

yes, but they have x^2 in the denom

Answer 73

yeah the integral shouldn't be problem

Answer 74

so that makes it tan^2

Answer 75

thanks again Turing :)

Answer 76

welcome :)