Question

(50 PONTOS) Calcule a integral indefinida:

$\displaystyle\int\!\sqrt{\sec^4 x+1}\,\cdot \mathrm{tg\,}x\,dx$

Answer 1

$\displaystyle\int\sqrt{\sec^4 x+1}\,\cdot \mathrm{\tan\,}x\,dx\\\\ x=\arccos u\\\\ \displaystyle\int\sqrt{\sec^4(\arccos u)+1}.\tan (\arccos u)dx\\\\ \displaystyle\frac{dx}{du}=-\displaystyle\frac{1}{\sqrt{1-u^2}}\implies dx=-\displaystyle\frac{1}{\sqrt{1-u^2}}du\\\\ \displaystyle\int\sqrt{\sec^4(\arccos u)+1}.\tan (\arccos u).-\displaystyle\frac{1}{\sqrt{1-u^2}}du\implies\\\\ \displaystyle\int-\frac{\sqrt{\frac{1}{u^4}+1}}{u}du\implies -\displaystyle\int\frac{\sqrt{\frac{1}{u^4}+1}}{u}du$
fazer substituição de novo:
$v=u^4$
de modo que
$\displaystyle\frac{dv}{du}=4u^3\implies du=\frac{dv}{4u^3}$
$\displaystyle-\int\frac{\sqrt{\frac{1}{v}+1}}{u}\frac{dv}{4u^3}=-\int\frac{\sqrt{\frac{1}{v}+1}}{4u^4}dv\ \ |v=u^4\implies\\ -\int\frac{\sqrt{\frac{1}{v}+1}}{4v}dv=-\frac{1}{4}\int\frac{\sqrt{\frac{1}{v}+1}}{v}dv$
aplicar substituição de novo:
$\displaystyle w=\frac{1}{v}\implies\frac{dw}{dv}=-\frac{1}{v^2}dv$
obtendo:
$\displaystyle-\frac{1}{4}\int -\frac{1}{w}\sqrt{w+1}dw\implies-\frac{1}{4}.-\int\frac{1}{w}\sqrt{w+1}dw$
substituição de novo:
$\displaystyle t=\sqrt{w+1}\implies \frac{dt}{dw}=\frac{1}{2t}\implies dw=2tdt$
então:
$\displaystyle-\frac{1}{4}\big(-\int\frac{t}{w}dt\big)=\displaystyle-\frac{1}{4}\big(-\int\frac{t}{w}2tdt\big)$
como $t=\sqrt{w+1}$ o nosso $w=t^2-1$
$\displaystyle-\frac{1}{4}\big(-\int\frac{2t^2}{(t^2-1)}dt\big)\implies -\frac{1}{4}\big(-2\int\frac{t^2}{(t^2-1)}dt\big) \implies\\\\ -\frac{1}{4}\big(-2\big(-\int\frac{t^2}{(t^2-1)}dt\big)\big) \implies -\frac{1}{4}\big(-2\big(-\int\frac{t^2+(-t^2+1)}{-t^2+1}dt\big)\big)\\\\ -\frac{1}{4}\big(-2\big(-\int\frac{t^2+(-t^2+1)}{-t^2+1}dt-\int dt\big)\big)$
$\displaystyle \frac{t^2-t^2+1}{-t^2+1}=\frac{1}{1-t^2}\\\\ \int\frac{1}{1-t^2}dt=arctanh(t)\\\\ \int dt=t$
(integral notável a do aarctanh)
$\displaystyle-\frac{1}{4}\big -2\big(-(arctanh(t)-t\big) \big)\\\\ t=\sqrt{w+1},\ w=\frac{1}{v},\ v=u^4,\ u=\cos x\\\\ \frac{1}{2}\big(-\big(arctanh\big( \sqrt{\sqrt{\frac{1}{\cos^4x}}+1}-\sqrt{\frac{1}{\cos^4x}+1}\big)\big)+C\implies \\\\ \underline{\frac{1}{2}\big (\sqrt{\sec^4x+1}-arctanh(\sqrt{\sec^4x+1})+C}$

ou seja:
$\displaystyle\int \sqrt{\sec^4x+1}\tan xdx=\frac{1}{2}\big (\sqrt{\sec^4x+1}-arctanh(\sqrt{\sec^4x+1})+C$

tg e sec(arccos(x)):
secante de arco-cosseno:
$\displaystyle \sec (x)=\frac{1}{\cos(x)}\implies \sec^4(\arccos(u))=\frac{1}{\cos^4(\arccos(u))}\\\\ \cos(\arccos(u))=u\implies \sec^4(arccos(u))=\frac{1}{u^4}$

tangente de arco-cosseno:
$\displaystyle \tan(x)=\frac{\sin(x)}{\cos(x)}\implies\tan(\arccos(x))=\frac{\sin(\arccos(x))}{\cos(\arccos(x))}\implies\\\\ \tan(\arccos(x))=\frac{\sin(arccos(x))}{x}\ |\sin(x)=\sqrt{\cos(x)-1}\ |\\\\ \tan(\arccos(x))=\frac{\sqrt{x-1}}{x}$
depois efetua a multiplicação.

Answer 2

$\int\limits^{}_{} { \sqrt{sec^4x+1} .tgx }\, dx$

Multiplicando e dividindo tgx por $4sec^4x$

$\int\limits^{}_{} { \sqrt{sec^4x+1} . \frac{4sec^4xtgx}{4sec^4x} }\, dx$

Substituição

$u = sec^4x\\\\du=4tg(x)sec^4(x)dx$

$\frac{1}{4} \int\limits^{}_{} { \frac{ \sqrt{1+u} }{u} }\, du$

Pela tabela de integrais

$\int\limits^{}_{} { \frac{ \sqrt{a+bu} }{u} }\, du=2 \sqrt{a+bu}+a \int\limits^{}_{} { \frac{1}{u \sqrt{a+bu} } }\, du$

Mas, ainda pela tabela de integrais, temos que

$\int\limits^{}_{} { \frac{1}{u \sqrt{a+bu} } }\, du = \frac{1}{a}ln| \frac{ \sqrt{a+bu}- \sqrt{a} }{ \sqrt{a+bu}+ \sqrt{a} }$

Mas para esse último se e somente se, a>0.

a = 1, condição satisfeita
b = 1

$\dfrac{1}{4} \int\limits^{}_{} { \dfrac{ \sqrt{1+u} }{u}= \dfrac{1}{4}.2 \sqrt{1+u} + \dfrac{1}{4} .ln| \dfrac{ \sqrt{1+u}-1 }{ \sqrt{1+u}+1 }|$

$u = sec^4x$

$\sqrt{sec^4x+1} .tgx }\, dx= \dfrac{1}{2} \sqrt{1+sec^4x}+ \dfrac{1}{4}.ln| \dfrac{ \sqrt{1+sec^4x}-1 }{ \sqrt{1+sec^4x}+1 }| + C$