use the indicated substitution to evaluate the integral. let (x = 14\tan(\theta)). (give an exact answer…

Explanation:

Step1: Find $dx$

If $x = 14\tan(\theta)$, then $dx=14\sec^{2}(\theta)d\theta$. Also, $\sqrt{x^{2}+196}=\sqrt{196\tan^{2}(\theta)+ 196}=14\sec(\theta)$ and $x^{2}=196\tan^{2}(\theta)$.

Step2: Change the limits of integration

When $x = \frac{1}{2}$, $\frac{1}{2}=14\tan(\theta)\Rightarrow\tan(\theta)=\frac{1}{28}\Rightarrow\theta=\arctan(\frac{1}{28})$. When $x = 1$, $1 = 14\tan(\theta)\Rightarrow\tan(\theta)=\frac{1}{14}\Rightarrow\theta=\arctan(\frac{1}{14})$.

Step3: Rewrite the integral

The integral $\int_{\frac{1}{2}}^{1}\frac{dx}{x^{2}\sqrt{x^{2}+196}}$ becomes $\int_{\arctan(\frac{1}{28})}^{\arctan(\frac{1}{14})}\frac{14\sec^{2}(\theta)d\theta}{196\tan^{2}(\theta)\times14\sec(\theta)}=\frac{1}{196}\int_{\arctan(\frac{1}{28})}^{\arctan(\frac{1}{14})}\frac{\sec(\theta)}{\tan^{2}(\theta)}d\theta$. Since $\frac{\sec(\theta)}{\tan^{2}(\theta)}=\frac{\cos(\theta)}{\sin^{2}(\theta)}$, let $u = \sin(\theta)$, then $du=\cos(\theta)d\theta$.

Step4: Integrate with respect to $u$

The integral $\frac{1}{196}\int_{\arctan(\frac{1}{28})}^{\arctan(\frac{1}{14})}\frac{\cos(\theta)}{\sin^{2}(\theta)}d\theta=\frac{1}{196}\int_{u_1}^{u_2}\frac{du}{u^{2}}$, where $u_1=\sin(\arctan(\frac{1}{28}))=\frac{1}{\sqrt{1 + 28^{2}}}$ and $u_2=\sin(\arctan(\frac{1}{14}))=\frac{1}{\sqrt{1+14^{2}}}$. $\frac{1}{196}\int_{u_1}^{u_2}u^{-2}du=\frac{1}{196}\left[- \frac{1}{u}\right]_{u_1}^{u_2}=\frac{1}{196}\left(-\frac{1}{u_2}+\frac{1}{u_1}\right)$. $u_1=\frac{1}{\sqrt{1 + 28^{2}}}=\frac{1}{\sqrt{785}}$ and $u_2=\frac{1}{\sqrt{1+14^{2}}}=\frac{1}{\sqrt{197}}$. So the integral is $\frac{1}{196}(\sqrt{197}-\sqrt{785})$.

Answer:

$\frac{\sqrt{197}-\sqrt{785}}{196}$