Integral Trig Substitution Question

[TheRedDevil18]

I just have a few questions. When using a trig substitution does it have to be under a radical ?

eg, suppose I wanted to integrate (x²)/(x²-9), I used a trig substitution of x = 3sec(t) and got the wrong answer and so apparently I had to use partial fractions

 

[h6ss]

I just have a few questions. When using a trig substitution does it have to be under a radical ?

eg, suppose I wanted to integrate (x²)/(x²-9), I used a trig substitution of x = 3sec(t) and got the wrong answer and so apparently I had to use partial fractions

In the following integral,

Try using the following substitution:

What do we have?

 

[Mark44]

When using a trig substitution does it have to be under a radical ?

No.

eg, suppose I wanted to integrate (x2)/(x2-9), I used a trig substitution of x = 3sec(t) and got the wrong answer and so apparently I had to use partial fractions

Then I'm pretty sure you made a mistake. If you use partial fractions, you need to rewrite the integrand as a proper rational expression, which you can do by polynomial long division. (There's another way, as well.)
You can also do this by trig substitution. Without seeing your work, I can't say why you got a wrong answer.

 

[TheRedDevil18]

Ok, I'm going to skip some of the substituting parts to this,

3*integral sec³t/tan(t) dt
= (1+tan²t)/(tan(t)) * sec(t) dt

Integrating that, I get

ln(sin(t)) + sec(t)

Is it correct so far ?, I know the last thing I must do is sub the x values, but I just want to make sure if this part is correct

 

[Mark44]

Ok, I'm going to skip some of the substituting parts to this,

3*integral sec³t/tan(t) dt

This part looks OK. I'll check the following work in a bit.

 

[Mark44]

Ok, I'm going to skip some of the substituting parts to this,

3*integral sec³t/tan(t) dt
= (1+tan²t)/(tan(t)) * sec(t) dt

You lost the factor of 3, and have omitted the integration sign.
BTW, it's easy enough to do in LaTeX - # # \int f(x)dx # # (omit the spaces between the pound signs).

Integrating that, I get

ln(sin(t)) + sec(t)

Nope. The factor of 3 is still missing on both terms. One of them should be 3 sec(t), but the other is wrong.
I suspect that you did this:
$$\int \frac{dt}{sin(t)} = ln(sin(t))$$
... which is incorrect. This is a somewhat tricky integral. Rather than deriving it here, I would advise just looking it up in a table of integrals.

Is it correct so far ?, I know the last thing I must do is sub the x values, but I just want to make sure if this part is correct

 

[TheRedDevil18]

My integration was wrong. Turns out I had to use integration by parts for that trig integral with the sec^3 and tan and it was very messy. I did the same problem using partial fractions and it was much easier to integrate

Thanks for the help, guys :)

 

[Mark44]

You don't need integration by parts. Starting from what you had in post #4
In these integrals, sec(t) = x/3.
$$\int \frac{3sec^3(t)}{tan(t)}dt$$
$$=3\int \frac{dt}{cos^2(t) sin(t)}$$
$$= 3\int sec^2(t)csc(t)dt$$
$$=3\int \frac{sec(t)(tan^2(t) + 1)dt}{tan(t)}$$
$$=3\int sec(t)tan(t)dt + 3 \int \frac{sec(t)dt}{tan(t)}$$
The first integral is very easy, and the second integral simplifies to $3\int csc(t)dt$ which isn't too bad if you look it up in a table.

After undoing the substitution, I get the same result as I got using partial fractions, after a bit of manipulation.