Question

How can I calculate this integral?

Answer 1

\[\int\limits(y^{2}+A) ^{1.5}dy\]
Thanks you!

Answer 2

Excuse me, it's -1.5 instead of 1.5

Answer 3

if A is constant, then substitute
y=tan\(\theta\)

Answer 4

is A constant, or is it there to denote something else?
(like dA, we have in double integrals or anything of the sort (?))

Answer 5

Oh sorry the substitution should be
\(y=\sqrt{~A}\tan\theta\)

Answer 6

No, A is constant

Answer 7

so:
\[dy=\sqrt{A}(1+(\tan \theta) ^{2})\]
is it correct?

Answer 8

leave dy as sec^2theta

Answer 9

d(theta) must be added at the end of it..

Answer 10

Well I mean
√A • sec²(theta) d(theta)

Answer 11

I meant no need rewriting the sec²(theta)

Answer 12

@SolomonZelman could u wait a few minutes so I have time to do it and send you the results?

Answer 13

The idea behind this substitution is the rule,

\(\color{#000000 }{ \displaystyle \tan^2z+1=\sec^2z}\)

and by the same tocken,

\(\color{#000000 }{ \displaystyle b\tan^2z+b=b(\tan^2z+1)=b\sec^2z}\)

■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■
So, when you are dealing with;

\(\color{#000000 }{ \displaystyle \int \frac{1}{\sqrt{(y^2+A)^3{\color{white}{\large |}}} }dy}\)

you substitute;
\(\color{#000000 }{ \displaystyle y=\sqrt{A~}\tan\theta}\)
\(\color{#000000 }{ \displaystyle dy=\sqrt{A~}\sec^2\theta~d\theta}\)

you get,

\(\color{#000000 }{ \displaystyle \int \frac{1}{\sqrt{(~\left[\sqrt{A~}\tan\theta\right]^2+A)^3{\color{white}{\large |}}} }\left(\sqrt{A~}\sec^2\theta~d\theta\right)}\)

Answer 14

and on... applying this idea
(I won't do your work, don't worry, just pushing)

Answer 15

Thank you very much! :)

Answer 16

Anytime:)

Answer 17

By the way,
http://tutorial.math.lamar.edu/Classes/CalcII/TrigSubstitutions.aspx

(Good web resource fro trig sub, but look at the very bottom of the page, it gives you the important forms)

Answer 18

is it correct up to this point or no?

Answer 19

@SolomonZelman

Answer 20

and then it becomes:
\[\frac{ 1 }{ A }(tg \theta+C1)-\frac{ 1 }{ A }(\frac{ 1 }{ \sin \theta }+C2)\]
is it correct?

Answer 21

checking...

Answer 22

\(\color{#000000 }{ \displaystyle \int \frac{1}{\sqrt{(~\left[\sqrt{A~}\tan\theta\right]^2+A)^3{\color{white}{\large |}}} }\left(\sqrt{A~}\sec^2\theta~d\theta\right)}\)

\(\color{#000000 }{ \displaystyle \int \frac{\sqrt{A~}\sec^2\theta}{\sqrt{(A\tan^2\theta+A)^3{\color{white}{\large |}}} }~d\theta}\)

\(\color{#000000 }{ \displaystyle \frac{1}{A}\int \frac{\sec^2\theta}{\sqrt{(\sec^2\theta)^3} }~d\theta}\)

\(\color{#000000 }{ \displaystyle \frac{1}{A}\int \frac{\sec^2\theta}{\sec^3\theta }~d\theta}\)

\(\color{#000000 }{ \displaystyle \frac{1}{A}\int \cos \theta ~d\theta}\)

Answer 23

I am not very sure about what you got in the very last line on the right....

Answer 24

What did you do there to get the last line on the right?

Answer 25

Oh, DAMN!!!!!!
Damnnnnn!!!!!!!!
I can't believe i did that!

Answer 26

i think it's cuz it's 5:17 AM here after a hard day working....

Answer 27

oh, you factored it out, incorrectly, expanding, even though you had a -3/2 power?

Answer 28

Oh, please. Let it go! I just can't believe it!

Answer 29

lol \( ....\)
But, you see what I' did (right?), you now have a simple integral to deal with..

Answer 30

\(\color{#000000 }{ \displaystyle \frac{1}{A}\int \cos \theta ~d\theta}\)

Answer 31

yes
\[\frac{ -1 }{ A }\sin \theta\]

Answer 32

+C, and it is not negative

Answer 33

but you know what's the problem here?

Answer 34

What is it?

Answer 35

d/dx sin(x) = cos(x)

Consequentially,
\(\int\) cos(x) dx = sin(x) + C

Answer 36

(that's why sine is positive)

Answer 37

\[y=A tg \theta\]
then I need to rewrite sin with tg

Answer 38

really,
\(\color{#000000 }{ \displaystyle y=\sqrt{A}\tan \theta }\)

Answer 39

So you need to rearange it, that's all

Answer 40

It leads to a quadratic equation, doesn't it?

Answer 41

\(\color{#000000 }{ \displaystyle y=\sqrt{A~}\tan \theta }\)
\(\color{#000000 }{ \displaystyle y/\sqrt{A~}=\tan \theta }\)
\(\color{#000000 }{ \displaystyle \theta=\arctan\left(y/\sqrt{A~}\right) }\)

that is what you get
(yes, you can simplify this)
\(\color{#000000 }{ \displaystyle \frac{1}{A}\sin\left[\arctan\left(y/\sqrt{A~}\right)\right] }\)

Answer 42

you know that
\(\tan(\arctan z)=z\)

So, all you need to do, is to find the expression for sine, in terms of tangent.

Answer 43

(let me know if you need an aid with some trig properties)

Answer 44

I'll ask if i need some more help @SolomonZelman , Unfortunately I can't give you more than one medal. I hope other people who read this do that. Though I think it's not so important for you.

Actually, the last time i solved such questions was around 12-13 years ago. thank you very much for the time.
Thank you very much.

Answer 45

Anytime, it is really not a problem. (And yes, I don't care about medals)...

btw, if you finishing the problem at this very moment, have you found the expression for \(\sin x\) in terms of \(\tan x\)?

Answer 46

Start from the fact that

\(\color{#000000 }{ \displaystyle \tan x = \sin x / \cos x }\)
\(\color{#000000 }{ \displaystyle \tan x = \sin x \sec x }\)
\(\color{#000000 }{ \displaystyle \sin x=\tan x / \sec x=\tan x /\sqrt{\sec^2x} }\)

Answer 47

but, in your case \(x\), is that blony,
\(\arctan\left(y/\sqrt{A~}\right)\)

(and remember that you know that \(\tan(\arctan z)=z\) ...)

Answer 48

I have to check if it's possible based on changes of variable I'd made before reaching to this point or not.
It's become a hard problem for me. Though it's not a homework or anything like that, it's made my brain so busy. And in such situations I cannot sleep, eat or do any other thing. I have to go to the end.
I'll have difficulty with determining C in integrals.

Answer 49

You can't find "C", unless you are given some derivative f'(x), and some point on f(x), or something of this sort.

Answer 50

I know. I have to solve the problem from the beginning again. It was a definite integral. So I need to determine the limits after each change of variable again.

Answer 51

Say I had the following problem

\(\color{#000000 }{ \displaystyle \int_{2}^{5}2xe^{x^2+3}dx }\)

I can see the u-substitution here;
\(\color{#000000 }{ \displaystyle u=x^2+3 }\)
\(\color{#000000 }{ \displaystyle du=2x~dx }\)

And the limits integrate change as follows;

Old New
limits limits
(with "x") (with "u")

\(\color{#000000 }{ \displaystyle x=\color{blue}{2}~~\quad \Longrightarrow \quad u=\color{blue}{2}^2+3~~\quad \Longrightarrow \quad ~~u=4+3=7 }\)
\(\color{#000000 }{ \displaystyle x=\color{blue}{5}~~\quad \Longrightarrow \quad u=\color{blue}{5}^2+3~~\quad \Longrightarrow \quad~~ u=25+3=28 }\)

So my new integral is;

\(\color{#000000 }{ \displaystyle \int_{7}^{28}e^{u}du }\)

Answer 52

Yes, Thank you very much for your complementary explanation. Actually I know how to do it. But in the case I have a problem It'd be kind if you guide me.

Answer 53

For all math that I know, I am happy to help (when I am online).

Answer 54

sorry, I'm not so good at English, sometimes I might make some mistakes.
Thanks again for your big big help. :)

Answer 55

I wouldn't say my English is the best. English is my third language, although you can't hear my accent :)

Answer 56

ahahaha, who guess what? maybe one day we speak. By the way! My accent seems very good. I guess it's better than your skills in Math... hahahaha. (juz kidding)

Ok, I don't wanna take your time more than this.
See you,

Answer 57

Have a good morning!

Answer 58

Cu