i need help, trignometric substitutions problem

[afcwestwarrior]

1. The problem statement, all variables and given/known data
∫1/ ((t^3) sqrt t^2-1) * dt


2. Relevant equations
sqrt x^2 - a^2 , x=a sec theta , substitution 0 less than or equal to theta less than pi/2
identity, sec^2 theta-1 = tan^2 theta


3. The attempt at a solution
ok here we go
t= sec theta , dt= sec theta * tan theta
sqrt t^2-1= sqrt sec^2 theta -1 = tan theta

so i plug it in now
∫1/ (t^3) sqrt t^2-1 * dt = ∫dt/ (t^3) sqrt t^2-1=∫ sec theta * tan theta/ sec^3 theta * tan theta

=∫(1/sec^2 theta) * d theta

what do i do now

do i put in 1+ tan^2x for sec^2 theta
and use the u substitution

[statdad]

If you have


\int \frac{1}{\sec^2 \theta} \, d\theta}


what do you know about the secant function in terms of something other than tangent?

[afcwestwarrior]

do i turn it into cos^2 theta

[statdad]

Yes -


\int \frac 1 {\sec^2 \theta} \, d\theta = \int \cos^2 \theta \, d\theta


What can be done with \cos^2 \theta ?

[afcwestwarrior]

it'll equal 1/2 (1+2cos) theta right

[statdad]

I'm having a difficult time parsing your final comment, but if you wrote


\cos^2 \theta = \frac 1 2 \left( 1 + 2 \cos \theta \right)


I would advise you to check the double-angle formula again (check what the
statement above says for \theta = 0 if you don't see why it is incorrect.) If I misinterpreted your work I apologize.

[afcwestwarrior]

That's what I meant. It's ok apology accepted.

[tiny-tim]

That's what I meant. It's ok apology accepted.

erm … statdad was tactfully apologising for suggesting that you'd got it wrong … which you have …

\cos^2 \theta = \frac 1 2 \left( 1 + 2 \cos \theta \right) is wrong.

Try again! :smile:

[statdad]

As TinyTim points out (thanks, by the way :smile:)


\cos^2 \theta = \frac 1 2 \left( 1 + 2 \cos \theta \right)


is not correct .
Look very carefully at your reference book and compare the version in it to the one you wrote and we've reposted.

[afcwestwarrior]

Ok thank you.