1. ## Trig Substitution Integral

I'm really spinning my wheels on this one:

$\displaystyle \int^2_{\sqrt{2}} \frac{1}{t^3\sqrt{t^2-1}}dt$

let $\displaystyle t = \sec \theta$ then $\displaystyle dt = \sec\theta \tan\theta$

I know that substituting $\displaystyle \sec\theta$ for $\displaystyle \sqrt{t^2-1}$ yields $\displaystyle \tan\theta$. This is what I get next (general antiderivative):

$\displaystyle \int \frac{\sec\theta \tan\theta d\theta}{\sec^3\theta \tan\theta}$

Assuming that this substitution is correct, I then simplify the equation by canceling common factors, and I get:

$\displaystyle \int\cos^2 \theta d\theta$

which integrates to $\displaystyle \frac{1}{2}\sin^2\theta + C$

So, two questions:

1) Am I right to this point (solving for the general antiderivative), and if not, where am I going wrong?

2) Where I am really stuck is in trying to get the definite integral. How do I convert my boundaries from t to $\displaystyle \theta$ or get my general antiderivative back into t? This is really confusing to me.

Can anybody help?

Thanks!

2. ## Re: Trig Substitution Integral

Originally Posted by joatmon
I'm really spinning my wheels on this one:

$\displaystyle \int^2_{\sqrt{2}} \frac{1}{t^3\sqrt{t^2-1}}dt$

let $\displaystyle t = \sec \theta$ then $\displaystyle dt = \sec\theta \tan\theta$

I know that substituting $\displaystyle \sec\theta$ for $\displaystyle \sqrt{t^2-1}$ yields $\displaystyle \tan\theta$. This is what I get next (general antiderivative):

$\displaystyle \int \frac{\sec\theta \tan\theta d\theta}{\sec^3\theta \tan\theta}$

Assuming that this substitution is correct, I then simplify the equation by canceling common factors, and I get:

$\displaystyle \int\cos^2 \theta d\theta$
Correct so far. However, your next step is incorrect. Check that again.

which integrates to $\displaystyle \frac{1}{2}\sin^2\theta + C$

So, two questions:

1) Am I right to this point (solving for the general antiderivative), and if not, where am I going wrong?

2) Where I am really stuck is in trying to get the definite integral. How do I convert my boundaries from t to $\displaystyle \theta$ or get my general antiderivative back into t? This is really confusing to me.

Can anybody help?

Thanks!
Once you finish correcting the above, I would draw a right triangle with theta as the angle, and the sides labeled appropriately for your trig substitution. Then you can figure out how to get back to the t domain. Alternatively, you can transform the limits into the theta domain. Either way should work.

3. ## Re: Trig Substitution Integral

try the substitution

(t^2) - 1 = u^2

4. ## Re: Trig Substitution Integral

Originally Posted by joatmon
I'm really spinning my wheels on this one:

$\displaystyle \int^2_{\sqrt{2}} \frac{1}{t^3\sqrt{t^2-1}}dt$

let $\displaystyle t = \sec \theta$ then $\displaystyle dt = \sec\theta \tan\theta$

I know that substituting $\displaystyle \sec\theta$ for $\displaystyle \sqrt{t^2-1}$ yields $\displaystyle \tan\theta$. This is what I get next (general antiderivative):

$\displaystyle \int \frac{\sec\theta \tan\theta d\theta}{\sec^3\theta \tan\theta}$

Assuming that this substitution is correct, I then simplify the equation by canceling common factors, and I get:

$\displaystyle \int\cos^2 \theta d\theta$

which integrates to $\displaystyle \frac{1}{2}\sin^2\theta + C$

So, two questions:

1) Am I right to this point (solving for the general antiderivative), and if not, where am I going wrong?

2) Where I am really stuck is in trying to get the definite integral. How do I convert my boundaries from t to $\displaystyle \theta$ or get my general antiderivative back into t? This is really confusing to me.

Can anybody help?

Thanks!
A hyperbolic substitution might be more appropriate here, as $\displaystyle \displaystyle \sinh^2{x} \equiv \cosh^2{x} - 1$.

So in your case, make the substitution $\displaystyle \displaystyle t = \cosh{x} \implies dt = \sinh{x}\,dx$ and the integral $\displaystyle \displaystyle \int{\frac{dt}{t^3\sqrt{t^2 - 1}}}$ becomes

\displaystyle \displaystyle \begin{align*} \int{\frac{\sinh{x}\,dx}{\cosh^3{x}\sqrt{\cosh^2{x } - 1}}} &= \int{\frac{\sinh{x}\,dx}{\cosh^3{x}\sinh{x}}} \\ &= \int{\frac{dx}{\cosh^3{x}}} \\ &= \int{\frac{\cosh{x}\,dx}{\cosh^4{x}}} \\ &= \int{\frac{\cosh{x}\,dx}{(1 + \sinh^2{x})^2}} \\ &= \int{\frac{du}{(1 + u)^2}}\textrm{ after making the substitution }u = \sinh{x} \implies du = \cosh{x}\,dx \\ &= -\frac{1}{1 + u} + C \\ &= -\frac{1}{1 + \sinh{x}} + C \\ &= -\frac{1}{1 + \sqrt{\cosh^2{x} - 1}} + C \\ &= -\frac{1}{1 + \sqrt{t^2 - 1}} + C\end{align*}

5. ## Re: Trig Substitution Integral

Oh yeah. That was only slightly stupid... What I really meant to say was this:

$\displaystyle \int \cos^2 \theta d\theta = \frac{x}{2} + \frac{\sin2x}{4} + C = \frac{x + \sin x \cos x}{2} + C$

The right triangle part is where I am still struggling. I drew the triangle with theta as the angle, the hypotenuse labeled t and the x axis labeled a. Thus, $\displaystyle \sec \theta = \frac{t}{a}$

When t = 2, $\displaystyle \sec \theta = ???$ This should be pretty obvious, but somehow I'm having a brain hemorrhage on this. Am I approaching this correctly? What am I missing?

Thanks again.

6. ## Re: Trig Substitution Integral

Originally Posted by joatmon
Oh yeah. That was only slightly stupid... What I really meant to say was this:

$\displaystyle \int \cos^2 \theta d\theta = \frac{x}{2} + \frac{\sin2x}{4} + C = \frac{x + \sin x \cos x}{2} + C$

The right triangle part is where I am still struggling. I drew the triangle with theta as the angle, the hypotenuse labeled t and the x axis labeled a. Thus, $\displaystyle \sec \theta = \frac{t}{a}$

When t = 2, $\displaystyle \sec \theta = ???$ This should be pretty obvious, but somehow I'm having a brain hemorrhage on this. Am I approaching this correctly? What am I missing?

Thanks again.
Good start on your triangle. I would let a = 1, since that's the substitution you used earlier. Use the Pythagorean Theorem to determine the vertical side of the triangle. What do you get?

7. ## Re: Trig Substitution Integral

I think the values of $\displaystyle \cos^{-1}{\left(\frac{1}{2}\right)}$ and $\displaystyle \cos^{-1}{\left(\frac{1}{\sqrt{2}\right)}$ are pretty standard.
That being said, deriving them from a sketch is probably a good practice.