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I have the following definite integral:

$$10\int_0^{\ln3} \frac{e^tdt}{\sqrt{e^{2t}+4}}$$

Using the following substitution:

$$e^t=2\tan\theta$$ $$dt=\frac{\sec^2\theta}{\tan\theta}d\theta$$

I get this:

$$2\int_{\theta_1}^{\theta_2}d\theta\frac{2\tan\theta}{\sqrt{4(1+\tan^2\theta)}}\cdot\frac{\sec^2\theta}{\tan\theta}$$

Simplifying:

$$2\int_{\theta_1}^{\theta_2}\frac{2\tan\theta\sec^2\theta}{2\tan\theta\sec\theta}d\theta$$

$$2\int_{\theta_1}^{\theta_2}\sec\theta d\theta$$

$$2\ln|\sec\theta+\tan\theta|_{\theta_1}^{\theta_2}$$

Back-substituting:

$$2\ln|\frac{\sqrt{4+e^{2t}}+{e^t}}{2}|_0^{\ln3}$$

$$2\ln|\frac{\sqrt{13}+{3}}{2}|$$

I've made a mistake somewhere but can't figure out what is is. Some help would be much appreciated.

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  • $\begingroup$ At the beginning, I would've used the substitution $e^t=u\dots$ and it should be $dt=\frac{\sec^2\theta}{\color{red}2\tan\theta}d\theta$ $\endgroup$ Oct 1, 2016 at 22:59
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    $\begingroup$ The coefficient in front of the integral should be 10 instead of 2, and you neglected to subtract the value of the integrand when $t=0$ $\endgroup$
    – user84413
    Oct 2, 2016 at 0:35

1 Answer 1

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Letting $u=e^t, du=e^tdt$ gives $\;\displaystyle10\int_0^{\ln3} \frac{e^tdt}{\sqrt{e^{2t}+4}}=10\int_1^3\frac{1}{\sqrt{u^2+4}}du$.

Now let $u=2\tan\theta, du=2\sec^2\theta$ to get

$10\displaystyle\int_{u=1}^{u=3}\frac{1}{2\sec\theta}\cdot2\sec^2\theta d\theta=10\int_{u=1}^{u=3}\sec\theta d\theta=10\left[\ln\big|\sec\theta+\tan\theta\big|\right]_{u=1}^{u=3}$

$\displaystyle=10\left[\ln\bigg|\frac{\sqrt{u^2+4}}{2}+\frac{u}{2}\bigg|\right]_1^{3}=10\left[\ln(\sqrt{u^2+4}+u)\right]_1^3=10\left(\ln(\sqrt{13}+3)-\ln(\sqrt{5}+1)\right)$

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