1. ## Trig substitution

I've tried to substitute x in but seem to get the integration wrong.

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

2. $\displaystyle \int \frac{1}{x^2\sqrt{25-x^2}}dx$ let$\displaystyle x=5sin\theta$

2. Your choice of substitution is good. Don't forget dx. That can be overlooked

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

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

$\displaystyle 625\int{sin^{2}{\theta}cos^{2}{\theta}d{\theta}$

3. 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).

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

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

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

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

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

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

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

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

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

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

I need a nap . . .

4. Originally Posted by c_323_h
$\displaystyle \int \frac{1}{x^2\sqrt{25-x^2}}dx$
I do these problems in a formal way.
Ignore this, if you do not understand I do not want to confuse you.
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Let,
$\displaystyle u=\sin^{-1}(5x)$
Then,
$\displaystyle \frac{du}{dx}=\frac{1}{\sqrt{25-x^2}}$--->Denominator.
Express integrand as,
$\displaystyle \int \frac{1}{x^2}\cdot \frac{1}{\sqrt{25-x^2}}dx$
Also, we have,
$\displaystyle \sin u=5x$
Thus,
$\displaystyle \csc u=\frac{1}{5x}$
Thus,
$\displaystyle \csc^2 u=\frac{1}{25x^2}$
Thus,
$\displaystyle \frac{1}{x^2}=25\csc^2 u$
Thus, you finally have,
$\displaystyle \int 25\csc^2u \frac{du}{dx}dx=25\int \csc^2udu=-25\cot u+C$
Substitute back,
$\displaystyle -25\cot (\sin^{-1}(5x))+C$
Simplify the inverse trigonometric with trigonometric, to get,
$\displaystyle -25\cdot \frac{\sqrt{1-25x^2}}{5x}+C$
Thus,
$\displaystyle -\frac{5\sqrt{1-25x^2}}{x}+C$