Question

Integral 2dx/ (x^2*(x^2+9)^(1/2)) ?

Answer 1

Yesterday you were doing derivatives, now you're doing integrals?

Answer 2

yes, I tried to substitute x=3 tan theta, but then I got stuck later on..

Answer 3

So you have: \[
\frac{2}{9\tan^2\theta \cdot 3\sec\theta}
\]

Answer 4

there should be dx on top as well substituted with 2* 3 sec^2 theta d theta

Answer 5

Wait so we have: \[
\frac{6\sec\theta}{27\tan^2\theta}
\]

Answer 6

Multiply top and bottom by \(\cos^2\theta\):\[
\frac{2\cos\theta}{9\sin^2\theta}
\]

Answer 7

\[
\cos\theta d\theta = d(\sin\theta)
\]

Answer 8

Meaning that \[
\frac{2\cos\theta\;d\theta}{9\sin^2\theta} = \frac{2\;d(\sin\theta)}{9\sin^2\theta}= \frac{2\;du}{9u^2}
\]

Answer 9

I am a little confused how you got 6secθ/ 27tan2θ and then 2cosθ/ 9sin2θ

Answer 10

we got 2/ 9 tan ^2 theta * 3 sec theta.... did you expressed it in terms of sin and cos?

Answer 11

Yes

Answer 12

I got 2 cos^3 theta/ 27 sin^2 theta

Answer 13

\[
\int \frac{2dx}{x^2\sqrt{x^2+9}}
\]

Answer 14

\[
\begin{split}
\int \frac{2\;dx}{x^2\sqrt{x^2+9}}
&=
\int \frac{2\;d(3\tan\theta)}{(3\tan\theta)^2\sqrt{(3\tan\theta)^2+9}}\\
&=
\int \frac{6\sec^2\theta \;d\theta}{9\tan^2\theta|3\sec\theta|}\\
&=
\int \frac{2\sec\theta \;d\theta}{9\tan^2\theta} \times\frac{\cos^2\theta}{\cos^2\theta}\\
&=
\int \frac{2\cos\theta \;d\theta}{9\sin^2\theta} \\
&=
\int \frac{2\;d(\sin\theta)}{9(\sin\theta)^2} \\
&=
\int \frac{2\;du}{9u^2} \\
&=
\frac{2}{9}\int u^{-2}\;du \\
&=
\frac{2}{9}\int d\left(-u^{-1} \right)\\
&= -\frac{2}{9 u}+C
\end{split}
\]

Answer 15

So the final answer in terms of x would be -2(9+x^2)^(1/2) / 9x ??

Answer 16

Well, \[
u = \sin\theta\\
x=3\tan\theta
\]

Answer 17

\[
\theta=\tan^{-1}\left(\frac{x}{3}\right)
\]

Answer 18

\[
u = \sin\left(\tan^{-1}\left(\frac x3\right)\right)
\]

Answer 19

\[
u = \frac{x}{\sqrt{x^2+9}}
\]

Answer 20

So \[
-\frac {2}{9u}+C = -\frac{2\sqrt{x^2+9}}{9x}+C
\]

Answer 21

that's what I got as well ! Thank you for double checking!