T-formula integration (2 Viewers)

blackops23 · Jul 6, 2011

Hey guys, just wondering is there any way that you can use:

t=tan(x/2) in order to simply/easily integrate 1/(1+3cos2x)?

Or is it advisable to use t=tanx?

Thanks

hscishard · Jul 6, 2011

Try t=tanx, if that's not cool, expand the cos2x to it's trig(x) parts

blackops23 · Jul 6, 2011

hscishard said:
Try t=tanx, if that's not cool, expand the cos2x to it's trig(x) parts

t=tanx worked normally, but the question specified t=tan(x/2)

When I tried it, I used the double angle formula while resulted in a (cosx)^2 being in the expression, which resulted in ((1-t^2)/(1+t^2))^2 --. ugly and complicated stuff

anyone know if theres a quick easy way to use tan(x/2) = t, or is tanx= t the only proper way?

Hermes1 · Jul 6, 2011

hey blackops why are u doin so many integration questions, you should be focusing on harder topics like conics and mechanics for ur trial

Trebla · Jul 7, 2011

Did the question say something like "use t = tan (x/2)" or just "use t-formula substitution"?

blackops23 · Jul 7, 2011

Trebla said:
Did the question say something like "use t = tan (x/2)" or just "use t-formula substitution"?

it said use t=tan(x/2), but its a coroneos question, and i'd doubt it'd be in an exam,

Anyways, can someone please help me on this t-formula integration which involves trigonometric squares?

Q. INT (sinx^2)/(1+cosx) dx

I did it and ended up with having to integrate (2t^2)/(1+t^2)^2

Did I do something wrong, or is there a particular trick with this question?

As always, help is greatly appreciated

EDIT -- difference of two squares cancel out, dw guys...

Drongoski · Jul 7, 2011

blackops23 said:
it said use t=tan(x/2), but its a coroneos question, and i'd doubt it'd be in an exam,

Anyways, can someone please help me on this t-formula integration which involves trigonometric squares?

Q. INT (sinx^2)/(1+cosx) dx

I did it and ended up with having to integrate (2t^2)/(1+t^2)^2

Did I do something wrong, or is there a particular trick with this question?

As always, help is greatly appreciated

EDIT -- difference of two squares cancel out, dw guys...

You already got it:

$\int \frac {sin^2 x}{1+cos x} dx = \int \frac {1-cos^2 x}{1+cos x} dx = \int (1-cos x)dx = x - sin x + C$

For the t-formula better in this case to use: x = tanx

Did it last nite - but don't know where I threw away soln. Have to re-do. In my hurry to post solution, it may well be wrong!

Let t = tan x

$\therefore x = tan ^{-1}t \implies dx = \frac {1}{1+ t^2}dt$

$cos x =\frac {1}{\sqrt {1+t^2}}$ and $sin x = \frac {t^2}{\sqrt{1+t^2}}$ and $cos2x = cos^2 x-sin^2 x = \frac {1-t^2}{1+t^2}$

$\therefore \int \frac {1}{1+3cos2x} dx = \int \frac {1}{1+3 \times \frac {1-t^2}{1+t^2}} \cdot \frac {1}{1+t^2}dt = \int \frac {1}{4-2t^2} dt$

$= \int \frac {1}{4} \left [\frac {1}{2+\sqrt 2 t} + \frac {1}{2- \sqrt 2 t} \right ]dt = \frac {1}{4} \left [ \frac {1}{\sqrt 2} ln \left | \frac {2+\sqrt2 t}{2- \sqrt 2 t}\right |\right ] + C$

$= \frac {1}{4 \sqrt 2 } ln \left| \frac {2 + \sqrt 2 tan x}{2 - \sqrt 2 tan x} \right | + C$

limit1 · Jul 7, 2011

you can do it by complex analysis newbs

pretty easy way too

blackops23 · Jul 7, 2011

Lol jeez Drongos, I already got the answer.. you didn't have to waste time with that massive solution lol...

and i know this is the billionth time I've asked for help, but could someone please outline how to do this:

Q. INT (1+cosx)/(1+sinx) dx

I used t=tan(x/2)

and after simplification, ended up having to integrate:

4*INT1/[(1+t^2)(t^2 + 2t+ 1)]

I really don't want to use partial fractions, and with all of these questions, there is always a faster trick.

Any ideas guys?

Help is greatly appreciated

Drongoski · Jul 7, 2011

How about?

$= \int \frac {(1+cos x)(1-sin x)}{(1+sin x)(1-sin x)} dx = \int \frac{1-sin x + cos x - sinx cosx }{cos ^2 x } dx$

$= \int [sec^2 x -secx tanx + sec x - tanx]dx$

all of which can be readily done.

blackops23 · Jul 7, 2011

Drongoski said:
How about?

$= \int \frac {(1+cos x)(1-sin x)}{(1+sin x)(1-sin x)} dx = \int \frac{1-sin x + cos x - sinx cosx }{cos ^2 x } dx$

$= \int [sec^2 x -secx tanx + sec x - tanx]dx$

yeah what the other guy said, thankyou for the fast method! but do you see any way with using the t-formula?

Drongoski · Jul 7, 2011

Which other guy?

blackops23 · Jul 7, 2011

Drongoski said:
Which other guy?

I think he deleted his comment :S

Hermes1 · Jul 7, 2011

blackops23 said:
yeah what the other guy said, thankyou for the fast method! but do you see any way with using the t-formula?

with t method u have to use partial fractions at the end theres no other way

blackops23 · Jul 7, 2011

can someone confirm whether this is true?

can t-formula only be used on integrals in the form of dx/(a+bcosx + csinx)?

EDIT: for the question, integrate (1+cosx)/(1+sinx), the answer was:

-2/[1+tan(x/2)] + ln(1+sinx) + C

How did Coroneos achieve this answer?

Hermes1 · Jul 7, 2011

blackops23 said:
can someone confirm whether this is true?

can t-formula only be used on integrals in the form of dx/(a+bcosx + csinx)?

EDIT: for the question, integrate (1+cosx)/(1+sinx), the answer was:

-2/[1+tan(x/2)] + ln(1+sinx) + C

How did Coroneos achieve this answer?

i think what coroneos did was split it up into two integrals so 1/1+sin x and cos x/1+sin x

he then used t method on the first integral and the second one is just integration of ln (cos x is the derivative of 1 + sinx)

thats why he got that answer

i dont think t method is limited to integrals of that form, i think coroneos just used the fastest method of solving the question

cutemouse · Jul 7, 2011

arj1 said:
i think what coroneos did was split it up into two integrals so 1/1+sin x and cos x/1+sin x

he then used t method on the first integral and the second one is just integration of ln (cos x is the derivative of 1 + sinx)

thats why he got that answer

i dont think t method is limited to integrals of that form, i think coroneos just used the fastest method of solving the question

Probably...

Also, I think you can let u=2x and then proceed from there... But that's basically the same as using the subs t=tanx because then dx = dt/(1+t^2), whereas the the t=tan(x/2) subs yields dx = 2dt/(1+t^2)

blackops23 · Jul 7, 2011

arj1 said:
i think what coroneos did was split it up into two integrals so 1/1+sin x and cos x/1+sin x

he then used t method on the first integral and the second one is just integration of ln (cos x is the derivative of 1 + sinx)

thats why he got that answer

i dont think t method is limited to integrals of that form, i think coroneos just used the fastest method of solving the question

yep I did what you said and got the same answer, (but it was longer than Drongo's method)

Fuuarkk... is there any way to know the fastest integration method? Or is it just practice...?

Hermes1 · Jul 7, 2011

blackops23 said:
yep I did what you said and got the same answer, (but it was longer than Drongo's method)

Fuuarkk... is there any way to know the fastest integration method? Or is it just practice...?

its just practice. i time myself doing the 1st question of each past paper with integration an it usually takes 10 minutes max. you should be aiming to have questions 1 and 2 finished in about 15-20 minutes, bcuz curve sketching takes up a lot of time in exams.

blackops23 · Jul 7, 2011

arj1 said:
its just practice. i time myself doing the 1st question of each past paper with integration an it usually takes 10 minutes max. you should be aiming to have questions 1 and 2 finished in about 15-20 minutes, bcuz curve sketching takes up a lot of time in exams.

Q1 is a joke -- most of the time, but often you might get some harder integration in Q4/5...

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