Question

a serious question this time can someone give me a trigonometric substitution problem please? i'd like to practice...and for all those people who'll answer "google" or something...i need someone to check if i did it right :P and who else is better to check than the one who gave the problem himself

Answer 1

@lgbasallote nice but what do u want??

Answer 2

....a trigonometric substitution problem....

Answer 3

it's integral calculus...if that's what you're asking

Answer 4

ok i have been studying the in-depth processes of trigonometric substitution with integrals and this problem has me frusterated.
\[\int\limits_{}x^2\sqrt{(x^2-4)}dx \]
The evaluation is clear (from an old Table of Integrals I found), but the derivation is not at all clear, which is what i want to know.

I also tried to solve this by integration by parts, but every approach ended with an even more complicated integral, so trig substitution is probably the best choice.

Answer 5

Perhaps you can try these two? (well.. I haven't done that neither :P 0

Answer 6

Try finding the area of an ellipse with major axis a and minor axis b, if you don't know it yet.

Answer 7

ill do yours first @Callisto it's easier lol..well the second one is the same so ill do the first...

Answer 8

not that yet @blockcolder :P

Answer 9

But it's good practice! =))

Answer 10

\(\LARGE \int \frac{1}{1+ 4x^2} dx\)

let \(\tan \theta = 2x\)
\(\sec^2 \theta d\theta = 2 dx\)
\(\sec \theta = \sqrt{1 + 4x^2}\)
\(\sec^2 \theta = 1+ 4x^2 \)

substituting those....

\(\LARGE \frac{1}{2} \int \frac{\sec^2 \theta d\theta}{\sec^2 \theta} \)

\(\frac{1}{2} \int d\theta\)
\(\frac{1}{2} (\theta)\)

since \(\tan \theta = 2x\)
\(\theta = \tan^{-1} 2x\)

so...

\(\LARGE \frac{\tan^{-1} (2x)}{2}\)

Answer 11

that's number 1 @Callisto

Answer 12

@Rohangrr , The solution to your problem is add and subtract 4.


\[(x^2 - 4) \sqrt{x^2-4}+4\sqrt{x^2-4}\]

Can you do it from here or you want me to proceed??

Answer 13

nope @shivam_bhalla
but still tryin' i asked my teacher as well!

Answer 14

Yes, but you forgot to add '+C'. ( Just my opinion only. )

Answer 15

I meant the answer. ( you know I don't do well in integration!)

Answer 16

oh yeah that too.. I LOST MY LATEX ON THE SECOND PROBLEM T_T

Answer 17

to rohangrr's problem i go

Answer 18

first step...the triangle...


from that we have

\(2 \sec \theta = x\)
\(2 \tan \theta = \sqrt {x^2 - 4}\)
\(2 \sec \theta \tan \theta d\theta = dx\)

so...trig sub...

\(\int (2\sec \theta)^2 (2\tan \theta) (2 \sec \theta \tan \theta d \theta)\)
\( 16 \int (\sec^2 \theta) (\tan \theta) (\sec \theta \tan \theta d\theta)\)
\( 16 \int (\sec^2 \theta) (\tan ^2 \theta) (\sec \theta d\theta)\)
\( 16 \int (\sec^2 \theta) (\sec ^2 \theta - 1) (\sec \theta d\theta)\)
\( 16 \int (\sec^3 \theta) (\sec ^2 \theta - 1)d \theta\)
\( 16 \int (\sec ^5 \theta - \sec ^3 \theta)d \theta\)
that's integrable right @Rohangrr

Answer 19

im going to redo callisto's problem now...

Answer 20

Ok. I will complete it for you

\[\int(x^2-4)^{3/2} + 4 * \int\sqrt{x^2- 2^2}\]

From here I will do the integreal for each part separartely

Answer 21

Or look at @lgbasallote solution. It is better than what I have done here on paper :)

Answer 22

\(\int (y\sqrt {1 - y^2})dy\)

let \(\cos \theta = y\)
\(-\sin \theta d \theta = dy\)
\(\sin \theta = \sqrt{1 - y^2}\)

\(\int (\cos \theta)(\sin \theta)(-\sin \theta d\theta)\)
\(-\int (\cos \theta)(\sin^2 \theta)d\theta\)
\(-\int (\cos \theta)(1 - \cos^2 \theta)d\theta\)
\(-\int (\cos \theta - \cos^3 \theta)d\theta\)
\(-\int \cos \theta d\theta + \int \cos^3 \theta d\theta\)
\(-\sin \theta + \int (1 - \sin^2 \theta)(\cos \theta) d\theta\)

let \(\sin \theta =\ s\)
\(\cos \theta d\theta = ds\)

\(-\sin \theta + \int (1 - s^2)ds\)
\(\LARGE -\sin \theta + s - \frac{s^3}{3}\)

\(\large -\sin \theta + \sin \theta - \frac{\sin^3 \theta}{3}\)

\(\large \frac{\sin ^3 \theta}{3}\)

\(\Large \frac{(1 - y^2)^{\frac{3}{2}}}{3}\)

@Callisto

Answer 23

The answer to Callisto's second problem is

\[\int\limits_{}^{} \ln(1+4x^2) dx\]

Applying u dv rule which is
\[\int\limits_{}^{} u dv = u \int\limits_{}^{}dv + v \int\limits_{}^{}du \]

Here u=ln(1+4x^2)
dv = dx

Answer 24

\[I=\ln(1+4x^2)\int\limits_{}^{}dx + \int\limits_{}^{}x * (d/dx)(\ln(1+4x^2)\]

\[I=xln(1+4x^2)+ \int\limits_{}^{}8x^2 dx / (1+4x^2) \]

\[I=xln(1+4x^2) + \ln(1+4x^2) +C\]
\[I=(x+1)(\ln(1+4x^2)) +C\]

Answer 25

Soory It is (x-1)(ln (1+4x^2))+C

I made a sign mistake in second term of udv rule . It should be negative

Answer 26

I did it like this:
\[\int y\sqrt{1-y^2}dy \]
Let y=sinθ, dy = cosθ dθ
\[\int sinθ\sqrt{1-sin^2θ}cosθdθ = \int sinθcos^2θdθ = -\int cos^2θ d(cosθ)\]\[ = (-1/3) cos ^3 θ +C = [-(1-y^2)^{(3/2)}] \ /3 +C\]

Answer 27

Sorry, I type slow :S

Answer 28

lol i guess that was faster =))

Answer 29

i meant the method

Answer 30

You forgot a '-' sign for your answer :S

Answer 31

@Callisto , nobody can defeat me in typing slowly :P
BTW, your method was however faster :) and yes you forgota a negative sign :P

Answer 32

@lgbasallote Check your last 3 steps

Answer 33

oh yeah lol...it's just typing latex so long makes you skip those :p =)))

Answer 34

@shivam_bhalla Well... Now you know who can :)

Answer 35

Haha :P I think we give each other a close competition in typing you know :P

BTW, Do you want to solve Qn 5 part b) ?? @Callisto

Answer 36

\(\Huge \color{forestgreen}{\mathtt{modesty}}\) \(\Huge \color{skyblue}{\mathcal{WARS!!!}}\)

Answer 37

@shivam_bhalla That's easy :S

Answer 38

LOLOLOL.............

Answer 39

wow...and you said you werent good in integration :P

Answer 40

@lgbasallote Then, I lose :) Mine is not modesty, it is honesty :)

Answer 41

@lgbasallote I'm not... Certainly not!

Answer 42

(not good at doing integration problems!)

Answer 43

@lgbasallote you said in the beginning " this time <silence> "

LOL. We are violating these conditions to the core :P

Answer 44

ah we are opposite..im better in these problems lol

@shivam_bhalla lol it was a joke..serious -> silence haha

Answer 45

Ok. Any more tough problems?? Or I shall give tough problems??

Answer 46

i am assuming that question is not directed for me as i have not given any problem so i doubt i should give MORE

Answer 47

Do you really want to try something difficult?

Answer 48

Well this is for everyone :)

Evaluate the following expression :
\[\int\limits_{}^{}dx/(\sin^4 x+ \cos^4 x)\]

Answer 49

@Callisto , Yes

Answer 50

@shivam_bhalla that is trig sub right?

Answer 51

Well, not exactly. No trig sub

Answer 52

But yes you have to use trig identities to solve this :)

Answer 53

I'm not sure.. I seem to have done something like that this morning :|

Answer 54

Should I apply double angle formula here?

Answer 55

No. I will give you a tip. Multiply and divide by sec^4 (x)
and in numerator split
sec^4 (x) = sec^2 (x)* sec^2 (x)
=(1+tan^2 (x))*(sec^2 (x))

Now try to solve. @Callisto

Answer 56

tip 2 = Multiply the sec^4x in the denominator and try to bring it in only tan form i.e all the terms must be tan and not sec.

Then take tan x = t and try to solve :)

Answer 57

So many tips.... :S

Answer 58

It's going to be something ugly :S

Answer 59

It is just the trailer. :P The film is still remaining :P

Answer 60

I don't have a good feeling... (Shivering)

Answer 61

Be positive. Take it as a challenge to solve this :) Say it is possible. Definitely you will arrive at the solution