I , trignometric substitutions problem

[afcwestwarrior]

Homework Statement

∫1/ ((t^3) sqrt t^2-1) * dt

Homework 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

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!

 

[statdad]

As TinyTim points out (thanks, by the way )

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

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.