use an appropriate substitution and then a trigonometric substitution to evaluate\n int_{0}^{ln 4}…

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

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

Answer

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)