# Trig substitution

• Jul 23rd 2006, 10:00 AM
c_323_h
Trig substitution
I've tried to substitute x in but seem to get the integration wrong.

1. $\int \frac{x^3}{\sqrt{x^2+9}} dx$ let $x=3tan\theta$

2. $\int \frac{1}{x^2\sqrt{25-x^2}}dx$ let $x=5sin\theta$
• Jul 23rd 2006, 11:36 AM
galactus
Your choice of substitution is good. Don't forget dx. That can be overlooked

For instance, #2. $x=5sin{\theta}\;\ dx=5cos{\theta}d{\theta}$

$\int{(25sin^{2}{\theta})\sqrt{25(1-sin^{2}{\theta})}5cos{\theta}d{\theta}}$

$625\int{sin^{2}{\theta}cos^{2}{\theta}d{\theta}$
• Jul 23rd 2006, 01:10 PM
Soroban
Hello, c_323_h!

Don't be discouraged . . . the first one takes forever
. . and there are a dozen places to make errors (I've made them all).

Quote:

$1)\;\;\int \frac{x^3}{\sqrt{x^2+9}}\,dx\qquad\text{Let }x = 3\tan\theta$

Let: $x = 3\tan\theta\quad\Rightarrow\quad dx = 3\sec^2\!\theta\,d\theta$
. . and $\sqrt{x^2 + 9}\:=\:\sqrt{9\tan^2\!\theta + 9} \:=\:\sqrt{9(\tan^2\!\theta + 1)} \:=$ $\sqrt{9\sec^2\!\theta} \:= \:3\sec\theta$

Substitute: . $\int\frac{(3\tan\theta)^3}{3\sec\theta}(3\sec^2\! \theta\,d\theta) \;= \;27\!\int\tan^3\!\theta\sec\theta\,d\theta$

We have: . $27\!\int\tan^2\!\theta(\sec\theta\tan\theta\,d \theta) \;= \;27\!\int(\sec\theta - 1)(\sec\theta\tan\theta\,d\theta)$

Let: $u = \sec\theta\quad\Rightarrow\quad du = \sec\theta\tan\theta\,d\theta$

Substitute: . $27\!\int(u^2 - 1)\,du \;= \;27\left(\frac{u^3}{3} - u\right) + C \;= \;9u(u^2 - 3) + C$

Back-substitute
. . Since $u = \sec\theta$, we have: . $9\sec\theta(\sec^2\!\theta - 3) + C$

Back-substitute: Since $\tan\theta = \frac{x}{3}$, then $\sec\theta = \frac{\sqrt{x^2+9}}{3}$

We have: . $9\cdot\frac{\sqrt{x^2+9}}{3}\left[\left(\frac{\sqrt{x^2+9}}{3}\right)^2 - 3\right] + C \;=$ $3\sqrt{x^2 + 9}\left(\frac{x^2 + 9}{9} - 3\right) + C$

. . $= \;3\sqrt{x^2+9}\left(\frac{x^2+9-27}{9}\right) + C \;= \;\frac{1}{3}\sqrt{x^2+9}\left(x^2 - 18\right) + C$

I need a nap . . .

• Jul 23rd 2006, 03:42 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
$
\int \frac{1}{x^2\sqrt{25-x^2}}dx
$

I do these problems in a formal way.
Ignore this, if you do not understand I do not want to confuse you.
------
Let,
$u=\sin^{-1}(5x)$
Then,
$\frac{du}{dx}=\frac{1}{\sqrt{25-x^2}}$--->Denominator.
Express integrand as,
$\int \frac{1}{x^2}\cdot \frac{1}{\sqrt{25-x^2}}dx$
Also, we have,
$\sin u=5x$
Thus,
$\csc u=\frac{1}{5x}$
Thus,
$\csc^2 u=\frac{1}{25x^2}$
Thus,
$\frac{1}{x^2}=25\csc^2 u$
Thus, you finally have,
$\int 25\csc^2u \frac{du}{dx}dx=25\int \csc^2udu=-25\cot u+C$
Substitute back,
$-25\cot (\sin^{-1}(5x))+C$
Simplify the inverse trigonometric with trigonometric, to get,
$-25\cdot \frac{\sqrt{1-25x^2}}{5x}+C$
Thus,
$-\frac{5\sqrt{1-25x^2}}{x}+C$