Solusi Integral Akar tan x dx
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Untuk menyelesaikan $\int \sqrt{tan x} \quad dx$, kita gunakan subsitusi $\sqrt{tanx}=y$.
$tan x=y^2$
$sec^2 x dx=2y dy$
$dx=\frac{2y}{sec^2 x} dy$
$dx=\frac{2y}{1+tan^2 x} dy$
$dx=\frac{2y}{1+y^4} dy$
Sehingga:
$\int \sqrt{tan x} \quad dx$
$=\int y\frac{2y}{1+y^4} dy$
$=\int \frac{2y^2}{1+y^4} dy$
$=\int \frac{(y^2 +1)+(y^2-1}{y^4+1} dy$
$=\int \frac{y^2+1}{y^4+1} dy + \int \frac{y^2-1}{y^4+1} dy$
$=\int \frac{1+ \frac{1}{y^2}}{y^2+\frac{1}{y^2}} dy$
$+ \int \frac{1- \frac{1}{y^2}}{y^2+\frac{1}{y^2}} dy$
$=l_1 + l_2$
$l_1$ pake subsitusi $y-\frac{1}{y}=t$
$l_2$ pake subsitusi $y+\frac{1}{y^2}$
Silahkan untuk melanjutkannya!
$tan x=y^2$
$sec^2 x dx=2y dy$
$dx=\frac{2y}{sec^2 x} dy$
$dx=\frac{2y}{1+tan^2 x} dy$
$dx=\frac{2y}{1+y^4} dy$
Sehingga:
$\int \sqrt{tan x} \quad dx$
$=\int y\frac{2y}{1+y^4} dy$
$=\int \frac{2y^2}{1+y^4} dy$
$=\int \frac{(y^2 +1)+(y^2-1}{y^4+1} dy$
$=\int \frac{y^2+1}{y^4+1} dy + \int \frac{y^2-1}{y^4+1} dy$
$=\int \frac{1+ \frac{1}{y^2}}{y^2+\frac{1}{y^2}} dy$
$+ \int \frac{1- \frac{1}{y^2}}{y^2+\frac{1}{y^2}} dy$
$=l_1 + l_2$
$l_1$ pake subsitusi $y-\frac{1}{y}=t$
$l_2$ pake subsitusi $y+\frac{1}{y^2}$
Silahkan untuk melanjutkannya!
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