Question

show the integral in which the substitution x = 2/7 tan(t) transforms

i = ∫ dx / √(49x² + 4)

(use symbolic notation and fractions where needed.)

i = ∫ 1/7 sec(t)

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.)

i =

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 integrand

First, substitute $x$ into $\sqrt{49x^{2}+4}$. We have $49x^{2}+4 = 49\times(\frac{2}{7}\tan(t))^{2}+4=49\times\frac{4}{49}\tan^{2}(t)+4 = 4\tan^{2}(t)+4=4(\tan^{2}(t) + 1)$. Since $\tan^{2}(t)+1=\sec^{2}(t)$, then $\sqrt{49x^{2}+4}=2\sec(t)$.
The original integral $I=\int\frac{dx}{\sqrt{49x^{2}+4}}$ becomes $I=\int\frac{\frac{2}{7}\sec^{2}(t)dt}{2\sec(t)}=\int\frac{1}{7}\sec(t)dt$.

Step3: Evaluate the integral in terms of $t$

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

Answer:

$\frac{1}{7}\ln|\sec(t)+\tan(t)|+C$