Question

Calcule a integral por substituição trigonométrica
HELP???

Answer 1

Resposta:

∫ 1/√(1/16+x²)   dx

Fazendo a substituição

x=(1/4)*tg(u)  ==> dx=(1/4)* sec²(u) du

∫ { 1/√[1/16+(1/16) *tg²(u)] }  *     (1/4)* sec²(u)  du

∫ { 1/(1/4)*√[1+tg²(u)] }  *     (1/4)* sec²(u)  du

∫ { 1/√[1+(tg²(u)] }  *     sec²(u)  du

∫ { 1/√[cos²(u)/cos²(u)+sen²(u)/cos²(u)] }  *     sec²(u)  du

∫ { 1/√[1/cos²(u)] }  *     sec²(u)  du

∫ { 1/√[sec²(u)] }  *     sec²(u)  du

∫  1/sec (u)  *     sec²(u)  du

multiplique por (sec(u)+tan(u))/(sec(u)+tan(u))

∫   sec (u)*(sec(u)+tan(u))/(sec(u)+tan(u))   du

∫ (sec²(u)+sec(u) *tan(u))/(sec(u)+tan(u))   du

Substitua

s=tan(u) +sec(u) ==>ds=sec²(u)+tan(s)*sec(u) du

∫ (sec²(u)+sec(u) *tan(u))/s)   ds /(sec²(u)+tan(s)*sec(u) )

∫ 1/s   ds  = ln |s| + c

Como s=tan(u) +sec(u)

= ln |tan(u) +sec(u)| + c

Como x=(1/4)*tg(u)  ==> u =arctan(4x)

= ln |tan(arctan(4x)) +sec(arctan(4x))| + c  é a resposta

Answer 2

$\large\boxed{\begin{array}{l}\rm Quando~o~integrando~tem~a~forma\\\rm\sqrt{a^2+x^2},use~a~substituic_{\!\!,}\tilde ao~x=a~tg(\theta)\\\rm~onde~dx=a~sec^2(\theta)~d\theta\\\rm e~o~radicando~\sqrt{a^2+x^2}~se~torna~\bf a\,sec(\theta).\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm observe~a~figura~que~anexei.}\\\sf vamos~ usar~a~substituic_{\!\!,}\tilde ao~x=\dfrac{1}{4}tg(\theta)\\\sf de~modo~que~dx=\dfrac{1}{4}sec^2(\theta)~d\theta\\\sf e~o~radical~\sqrt{\dfrac{1}{16}+x^2}~se~torna~\dfrac{1}{4}sec(\theta)\\\sf Assim,\\\displaystyle\sf\int\dfrac{1}{\sqrt{\dfrac{1}{16}+x^2}}~dx=\int\dfrac{\diagup\!\!\!\frac{1}{4}~\backslash\!\!\!\!sec^2(\theta)}{\diagup\!\!\!\frac{1}{4}~\backslash\!\!\!\!sec(\theta)}~d\theta \end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\sf\int sec(\theta)~d\theta=\ell n|sec(\theta)+tg(\theta)|+C\\\underline{\rm usando~o~tri\hat angulo~auxiliar~temos\!:}\\\sf tg(\theta)=\dfrac{cat~op}{cat~adj}=\dfrac{x}{\frac{1}{4}}=4x\\\sf sec(\theta)=\dfrac{hip}{cat~adj}=\dfrac{\sqrt{\frac{1}{16}+x^2}}{\frac{1}{4}}=4\sqrt{\dfrac{1}{16}+x^2}\\\underline{\rm substituindo~temos\!:}\\\displaystyle\sf\int\dfrac{1}{\sqrt{\dfrac{1}{16}+x^2}}~dx=\ell n\bigg|4\bigg(x+\sqrt{\dfrac{1}{16}+x^2}\bigg)\bigg|+C\end{array}}$