Question

use an appropriate substitution and then a trigonometric substitution to evaluate the integral $int_{0}^{ln 4}\frac{e^{t}dt}{sqrt{e^{2t}+25}}$. which substitution transforms the given integral into one that can be evaluated directly? a. $e^{t}=5sec\theta$ b. $e^{t}=5sin\theta$ c. $e^{t}=5\tan\theta$

Explanation:

Step1: Recall the form of the denominator

The denominator is $\sqrt{e^{2t}+25}=\sqrt{(e^{t})^{2}+25}$. We know the identity $\tan^{2}\theta + 1=\sec^{2}\theta$. If we let $e^{t}=5\tan\theta$, then $(e^{t})^{2}+25 = 25\tan^{2}\theta+25=25(\tan^{2}\theta + 1)=25\sec^{2}\theta$, and $\sqrt{(e^{t})^{2}+25}=5\sec\theta$. Also, $e^{t}dt = 5\sec^{2}\theta d\theta$.

Step2: Analyze other substitution options

If $e^{t}=5\sec\theta$, then $(e^{t})^{2}-25 = 25\sec^{2}\theta - 25=25\tan^{2}\theta$ which is not suitable for the given denominator form. If $e^{t}=5\sin\theta$, then $25-(e^{t})^{2}=25 - 25\sin^{2}\theta=25\cos^{2}\theta$ which is also not suitable for the given denominator $\sqrt{(e^{t})^{2}+25}$.

Answer:

C. $e^{t}=5\tan\theta$