Question

use an appropriate substitution and then a trigonometric substitution to evaluate
int_{0}^{ln 4} \frac{e^{t}dt}{sqrt{e^{2t}+25}}
a. (e^{t}=5sec\theta)
b. (e^{t}=5sin\theta)
c. (e^{t}=5\tan\theta)
rewrite the given integral using this substitution.
int_{0}^{ln 4} \frac{e^{t}dt}{sqrt{e^{2t}+25}}=int_{square}^{square}square d\theta

Explanation:

Step1: First - level substitution

Let \(u = e^{t}\), then \(du=e^{t}dt\). When \(t = 0\), \(u = e^{0}=1\); when \(t=\ln 4\), \(u = e^{\ln 4}=4\). The integral \(\int_{0}^{\ln 4}\frac{e^{t}dt}{\sqrt{e^{2t}+25}}\) becomes \(\int_{1}^{4}\frac{du}{\sqrt{u^{2}+25}}\).

Step2: Trigonometric substitution

Since we have \(\int\frac{du}{\sqrt{u^{2}+25}}\), and we know the form \(\sqrt{a^{2}+x^{2}}\) (here \(a = 5\) and \(x = u\)). We use the substitution \(u = 5\tan\theta\), then \(du=5\sec^{2}\theta d\theta\), and \(\sqrt{u^{2}+25}=\sqrt{25\tan^{2}\theta + 25}=5\sec\theta\).

Step3: Rewrite the integral

Substituting \(u = 5\tan\theta\) and \(du = 5\sec^{2}\theta d\theta\) into \(\int\frac{du}{\sqrt{u^{2}+25}}\), we get \(\int\frac{5\sec^{2}\theta d\theta}{5\sec\theta}=\int\sec\theta d\theta\). Also, when \(u = 1\), \(1 = 5\tan\theta\), so \(\tan\theta=\frac{1}{5}\), \(\theta=\arctan(\frac{1}{5})\); when \(u = 4\), \(4 = 5\tan\theta\), so \(\tan\theta=\frac{4}{5}\), \(\theta=\arctan(\frac{4}{5})\). So \(\int_{1}^{4}\frac{du}{\sqrt{u^{2}+25}}=\int_{\arctan(\frac{1}{5})}^{\arctan(\frac{4}{5})}\sec\theta d\theta\).

Answer:

\(\int_{\arctan(\frac{1}{5})}^{\arctan(\frac{4}{5})}\sec\theta d\theta\)