show the integral in which the substitution $x = \\frac{2}{7}\\tan(t)$ transforms $i=int\\frac{dx}{sqrt{49x^{…

show the integral in which the substitution $x = \\frac{2}{7}\\tan(t)$ transforms $i=int\\frac{dx}{sqrt{49x^{2}+4}}$. (use symbolic notation and fractions where needed.) $i=int\\frac{1}{7}sec(t)dt$. evaluate the integral $i$ in terms of $t$. (use symbolic notation and fractions where needed. use $c$ for the arbitrary constant. absorb into $c$ as much as possible.) evaluate the integral $i$ in terms of $x$. (use symbolic notation and fractions where needed. use $c$ for the arbitrary constant. absorb into $c$ as much as possible.)

show the integral in which the substitution $x = \\frac{2}{7}\\tan(t)$ transforms $i=int\\frac{dx}{sqrt{49x^{2}+4}}$. (use symbolic notation and fractions where needed.) $i=int\\frac{1}{7}sec(t)dt$. evaluate the integral $i$ in terms of $t$. (use symbolic notation and fractions where needed. use $c$ for the arbitrary constant. absorb into $c$ as much as possible.) evaluate the integral $i$ in terms of $x$. (use symbolic notation and fractions where needed. use $c$ for the arbitrary constant. absorb into $c$ as much as possible.)

Answer

Explanation:

Step1: Find $dx$

Given $x = \frac{2}{7}\tan(t)$, then $dx=\frac{2}{7}\sec^{2}(t)dt$.

Step2: Substitute $x$ into the denominator

If $x = \frac{2}{7}\tan(t)$, then $49x^{2}+4=49\times(\frac{2}{7}\tan(t))^{2}+4=49\times\frac{4}{49}\tan^{2}(t)+4 = 4\tan^{2}(t)+4=4(\tan^{2}(t) + 1)$. Since $\tan^{2}(t)+1=\sec^{2}(t)$, the denominator $\sqrt{49x^{2}+4}=\sqrt{4\sec^{2}(t)} = 2\sec(t)$.

Step3: Rewrite the integral

Substitute $dx$ and the denominator into the original integral $I=\int\frac{dx}{\sqrt{49x^{2}+4}}$, we get $I=\int\frac{\frac{2}{7}\sec^{2}(t)dt}{2\sec(t)}=\int\frac{1}{7}\sec(t)dt$.

Step4: Evaluate the integral in terms of $t$

The antiderivative of $\sec(t)$ is $\ln|\sec(t)+\tan(t)|+C$. So $I=\frac{1}{7}\ln|\sec(t)+\tan(t)|+C$.

Step5: Express the result in terms of $x$

Since $x = \frac{2}{7}\tan(t)$, then $\tan(t)=\frac{7x}{2}$. And $\sec(t)=\sqrt{1 + \tan^{2}(t)}=\sqrt{1+\frac{49x^{2}}{4}}=\frac{\sqrt{49x^{2}+4}}{2}$. So $I=\frac{1}{7}\ln\left|\frac{\sqrt{49x^{2}+4}}{2}+\frac{7x}{2}\right|+C=\frac{1}{7}\ln|\sqrt{49x^{2}+4}+7x|-\frac{1}{7}\ln(2)+C$. Since $-\frac{1}{7}\ln(2)$ can be absorbed into the constant $C$, $I=\frac{1}{7}\ln|\sqrt{49x^{2}+4}+7x|+C$.

Answer:

$I=\int\frac{1}{7}\sec(t)dt$ $I=\frac{1}{7}\ln|\sec(t)+\tan(t)|+C$ $I=\frac{1}{7}\ln|\sqrt{49x^{2}+4}+7x|+C$