Question

Integral por substituição :

∫dx sobre
(cos²x √tgx-1 ) =

Answer 1

$\displaystyle\int\dfrac{1}{cos^{2}(x)\sqrt{tg(x)-1}}dx=\int\dfrac{1}{\sqrt{tg(x)-1}}\dfrac{1}{cos^{2}(x)}dx$

Note que

$\dfrac{1}{cos^{2}(x)}=\left(\dfrac{1}{cos(x)}\right)^{2}=(sec(x))^{2}=sec^{2}(x)$

Então:

$\displaystyle\int\dfrac{1}{cos^{2}(x)\sqrt{tg(x)-1}}dx=\dfrac{1}{\sqrt{tg(x)-1}}sec^{2}(x)dx$

Fazendo $u=tg(x)-1~~\longrightarrow~~du=sec^{2}(x)dx$

Então:

$\displaystyle\int\dfrac{1}{cos^{2}(x)\sqrt{tg(x)-1}}dx=\int\dfrac{1}{\sqrt{u}}du\\\\\\\int\dfrac{1}{cos^{2}(x)\sqrt{tg(x)-1}}dx=\int u^{-\frac{1}{2}}du\\\\\\\int\dfrac{1}{cos^{2}(x)\sqrt{tg(x)-1}}dx=\dfrac{u^{-\frac{1}{2}+1}}{-(\frac{1}{2})+1}+constante\\\\\\\int\dfrac{1}{cos^{2}(x)\sqrt{tg(x)-1}}dx=\dfrac{u^{\frac{1}{2}}}{(\frac{1}{2})}+constante\\\\\\\int\dfrac{1}{cos^{2}(x)\sqrt{tg(x)-1}}dx=2u^{\frac{1}{2}}+constante$

$\\\\\\\boxed{\boxed{\int\dfrac{1}{cos^{2}(x)\sqrt{tg(x)-1}}dx=2(tg(x)-1)^{\frac{1}{2}}+constante}}$

Answer 2

$\int\ {\frac{1}{cos^2x\sqrt{tanx-1}}} \, dx$

$\int\ {\frac{sec^2x}{\sqrt{tanx-1}}} \, dx$

$u=tanx-1$

$du=sec^2x\ dx$

$dx=\frac{du}{sec^2x}$

$\int\ {\frac{sec^2x}{sec^2x\sqrt{u}}} \, du$

$\int\ {\frac{1}{\sqrt{u}}} \, du$

$\int\ {u^{-\frac12}} \, du$

$2\sqrt{u}+C$

$\boxed{2\sqrt{tanx-1}+C}$