# 2 Integral questions

• May 30th 2006, 03:12 PM
nirva
2 Integral questions
1. $\int \frac {1} {t^2 \sqrt {t^2 + 1}} dt$

2. $\int^{\pi}_{\pi/2} \cos(x)^2 \sin(x) dx$
Is this one's result - $\frac {1} {2}$?
• May 30th 2006, 03:17 PM
ThePerfectHacker
Quote:

Originally Posted by nirva

2. $\int^{\pi}_{\pi/2} \cos(x)^2 \sin(x) dx$
Is this one's result - $\frac {1} {2}$?

Let,
$u=\cos x$ then, $\frac{du}{dx}=-\sin x$ Thus, you have,
$\int^{\pi}_{\pi/2} \cos^2 x \sin x dx$ becomes,
$-\int^{-1}_{0} u^2 \frac{du}{dx} dx$
Thus,
$\left \int_{-1}^0 u^2 du=\frac{1}{3}u^3 \right| ^0_{-1}$
• May 30th 2006, 03:30 PM
ThePerfectHacker
Quote:

Originally Posted by nirva
1. $\int \frac {1} {t^2 \sqrt {t^2 + 1}} dt$

This one is more complicated.
This involves trigonomteric substitution.

You have,
$\int \frac{1}{t^2\sqrt{1+t^2}}dt$
Rationalize,
$\int \frac{\sqrt{1+t^2}}{t^2(1+t^2)}dt$
Let,
$u=\tan^{-1}t$
Then,
$\frac{du}{dt}=\frac{1}{1+t^2}$
$\csc u=\frac{\sqrt{1+t^2}}{t}$
$\cot u=\frac{1}{t}$
Thus, express integrand as,
$\int \frac{\sqrt{1+t^2}}{t}\cdot \frac{1}{t} \cdot \frac{1}{1+t^2} dt$
Thus, substitute the previous equations,
$\int \csc u \cot u \frac{du}{dx} dx$
Thus,
$\int \csc u\cot u du$
Now, just determine this new integral.

Use trigonometric identities thus,
$\int \frac{1}{\sin u}\cdot \frac{\cos u}{\sin u} du$
Thus,
$\int \frac{\cos u}{\sin^2u}du$
Let, $x=\sin u$
Then, $\frac{dx}{du}=\cos u$
Express integrand as,
$\int \frac{1}{\sin^2u} \cdot \cos u du$
Make the above substitutions,
$\int \frac{1}{x^2} \frac{dx}{du} du$
Thus,
$\int x^{-2} dx=-\frac{1}{x}+C$
Substitute for $x$ to get,
$-\frac{1}{\sin u}+C$
Thus,
$-\csc u+C$
Substitute for $u$ to get,
$-\csc (\tan^{-1}(t))+C$
Thus,
$-\frac{\sqrt{1+t^2}}{t}+C$
• May 31st 2006, 12:15 AM
TD!
I find the same answer but I don't see why you do the rationalizing and rewriting if the substitution t = tan(u) works immediately.

$
\int {\frac{{dt}}{{t^2 \sqrt {1 + t^2 } }}} \mathop \to \limits_{dt = \sec ^2 u du}^{t = \tan u} \int {\frac{{\sec ^2 udu}}{{\tan ^2 u\sqrt {1 + \tan ^2 u} }}} = \int {\frac{{\cos u}}{{\sin ^2 u}}du}
$

$
\int {\frac{1}{{\sin ^2 u}}d\left( {\sin u} \right) = - \frac{1}{{\sin u}} + C \to } - \frac{1}{{\sin \left( {\arctan t} \right)}} + C
$

Which is the same as your last step, yeilding:

$
-\frac{\sqrt{1+t^2}}{t}+C
$
• May 31st 2006, 02:20 PM
ThePerfectHacker
Quote:

Originally Posted by TD!
I find the same answer but I don't see why you do the rationalizing and rewriting if the substitution t = tan(u) works immediately.

$
\int {\frac{{dt}}{{t^2 \sqrt {1 + t^2 } }}} \mathop \to \limits_{dt = \sec ^2 u du}^{t = \tan u} \int {\frac{{\sec ^2 udu}}{{\tan ^2 u\sqrt {1 + \tan ^2 u} }}} = \int {\frac{{\cos u}}{{\sin ^2 u}}du}
$

$
\int {\frac{1}{{\sin ^2 u}}d\left( {\sin u} \right) = - \frac{1}{{\sin u}} + C \to } - \frac{1}{{\sin \left( {\arctan t} \right)}} + C
$

Which is the same as your last step, yeilding:

$
-\frac{\sqrt{1+t^2}}{t}+C
$

Indeed I did see that! You method lacks mathematical beauty mine does not.
• May 31st 2006, 02:23 PM
TD!
My beaty lies in compactness :D
• Jun 1st 2006, 02:02 PM
topsquark
Quote:

Originally Posted by TD!
My beaty lies in compactness :D

(Sigh!) I wish I was compact. Then I might be beautiful.

-Dan
• Jun 1st 2006, 02:07 PM
TD!
And not only beautiful, also closed and bounded! :D