Question

Solve given definite integral.

${\int }_1^2 \frac{1}{x\sqrt{9-x^2}}dx$

Short answer

$\left[\left[\frac{1}{3}\ln \left(\left|\tan \left(\frac{1}{2}{\sin }^{-1}\left(\frac{2}{3}\right)\right)\right|\right]-\left[\frac{1}{3}\ln \left(\left|\tan \left(\frac{1}{2}{\sin }^{-1}\left(\frac{1}{3}\right)\right)\right|\right)\right]\right.\right.\text{.}$

Step-by-step solution

Step1. Given Information

The integral is as follows.

${\int }_1^2 \frac{1}{x\sqrt{9-x^2}}dx$

The objective is to solve the integral.

Step2. Assumptions

To solve the integral, let $x=3\sin \left(u\right)$

Now, derivation of the above equation is solved below

$$\begin{gathered} x=3\sin \left(u\right) \\ dx=3\cos \left(u\right)du \end{gathered}$$

Step3. Solution

Now, use the identity $-{\sin }^2\left(u\right)+1={\cos }^2\left(u\right)$

So,

$\begin{array}{r}\frac{1}{x\sqrt{9-x^2}}=\frac{1}{3\sqrt{-9{\sin }^2\left(u\right)+9}\sin \left(u\right)}\\ =\frac{1}{9\sqrt{-{\sin }^2\left(u\right)+1}\sin \left(u\right)}\\ =\frac{1}{9\sqrt{{\cos }^2\left(u\right)}\sin \left(u\right)}\\ =\frac{1}{9\cos \left(u\right)\sin \left(u\right)}\end{array}$

Step4. Solution

The integral is solved below.

$\begin{array}{r}\frac{1}{3}\int \frac{1}{\sin \left(u\right)}du\\ =\frac{1}{3}\int \left(\frac{1}{2\sin \left(\frac{u}{2}\right)\cos \left(\frac{u}{2}\right)}\right)du\end{array}$

$=\frac{1}{3}\int \left(\frac{1}{2\sin \left(v\right)\cos \left(v\right)}\right)dv\left[v=\frac{u}{2},dv=\frac{du}{2}\right]$

$=\frac{1}{3}\int \frac{{\sec }^2\left(v\right)}{\tan \left(v\right)}dv$

$=\frac{1}{3}\int \frac{1}{w}dw\left[w=\tan v,dw={\sec }^2\left(v\right)dv\right]$

$\begin{array}{r}=\frac{1}{3}\ln \left(\right|w\left|\right)\\ =\frac{1}{3}\ln \left(\right|\tan \left(v\right)\left|\right)=\frac{1}{3}\ln \left(\left|\tan \left(\frac{1}{2}{\sin }^{-1}\left(\frac{x}{3}\right)\right)\right|\right)\end{array}$

$=\frac{1}{3}\ln \left(\left|\tan \left(\frac{u}{2}\right)\right|\right)$

Step5. Solution

The definite integral is solved below.

${\int }_1^2 \frac{1}{x\sqrt{9-x^2}}dx$

$={\left[\frac{1}{3}\ln \left(\left|\tan \left(\frac{1}{2}{\sin }^{-1}\left(\frac{x}{3}\right)\right)\right|\right)\right]}_1^2$

$=\left[\frac{1}{3}\ln \left(\left|\tan \left(\frac{1}{2}{\sin }^{-1}\left(\frac{2}{3}\right)\right)\right|\right)\right]-\left[\frac{1}{3}\ln \left(\left|\tan \left(\frac{1}{2}{\sin }^{-1}\left(\frac{1}{3}\right)\right)\right|\right)\right]$

Therefore, the value is$\left[\left[\frac{1}{3}\ln \left(\left|\tan \left(\frac{1}{2}{\sin }^{-1}\left(\frac{2}{3}\right)\right)\right|\right]-\left[\frac{1}{3}\ln \left(\left|\tan \left(\frac{1}{2}{\sin }^{-1}\left(\frac{1}{3}\right)\right)\right|\right)\right]\right.\right.\text{.}$