Question

como resolver essa integral trigonométrica |√x²+16 dx

Answer 1

A resolucao esta em anexo.

Answer 2

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$\displaystyle\sf\int\sqrt{x^2+16}~dx\\\underline{\rm fac_{\!\!,}a}\\\sf x=4~tg~\theta\implies dx=4sec^2~\theta\\\sf x^2=16tg^2~\theta\\\sf\sqrt{16+x^2}=\sqrt{16+16tg^2~\theta}=\sqrt{16(1+tg^2~\theta)}\\\sf\sqrt{16+x^2}=\sqrt{16sec^2~\theta}=4sec~\theta\\\displaystyle\sf\int\sqrt{x^2+16}~dx=\int4sec~\theta~4sec^2\theta~d\theta=16\int sec^3~\theta~d\theta\\\displaystyle\sf\int sec^3~d\theta=\dfrac{1}{2}sec~\theta\cdot tg~\theta+\dfrac{1}{2}\ell n|sec~\theta+tg~\theta| +k$

$\displaystyle\sf16\int sec^3~\theta~d\theta=16\cdot\dfrac{1}{2}sec~\theta\cdot tg~\theta+16\cdot\dfrac{1}{2}\ell n|sec~\theta+tg~\theta|+k\\\sf =8sec~\theta\cdot tg~\theta+8\ell n|sec~\theta+tg~\theta|+k$

$\sf tg~\theta=\dfrac{x}{4}\\\sf sec~\theta=\dfrac{\sqrt{16+x^2}}{4}\\\displaystyle\sf\int\sqrt{x^2+16}~dx=8\cdot\dfrac{\sqrt{x^2+16}}{4}\cdot\dfrac{x}{4}++8\ell n\left|\dfrac{x+\sqrt{x^2+16}}{4}\right|\\\displaystyle\sf\int\sqrt{x^2+16}~dx=\dfrac{\sqrt{x^2+16}}{2}+8\ell n\left|\dfrac{x+\sqrt{x^2+16}}{4}\right|+k$