Soal Integral Legendaris Akar Kuadrat tan x dx

Banyak yang sudah membahas soal ini di Youtube walaupun orang mancanegara. Integral ini tidak dapat diselesaikan menggunakan metode substitusi! Saya akan menyelesaikan dengan cara manipulasi dan tentunya melibatkan identitas trigonometri. Cara yang digunakan adalah mengubah √tan x menjadi 1/2 (√tan x + √cot x + √tan x – √cot x) dan dijadikan seperti berikut:

\displaystyle \int \sqrt{\tan x}\: dx\\ =\int \frac{\sqrt{\tan x}+\sqrt{\cot x}+\sqrt{\tan x}-\sqrt{\cot x}}{2}\: dx\\ =\int \left ( \frac{\sqrt{\tan x}+\sqrt{\cot x}}{2}+\frac{\sqrt{\tan x}-\sqrt{\cot x}}{2} \right )dx\\ =\frac{1}{2}\int \left ( \sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}} \right )dx+\frac{1}{2}\int \left ( \sqrt{\frac{\sin x}{\cos x}}-\sqrt{\frac{\cos x}{\sin x}} \right )dx

Kemudian kita gunakan manipulasi dengan cara mengkuadratkan \displaystyle \sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}} dan \displaystyle \sqrt{\frac{\sin x}{\cos x}}-\sqrt{\frac{\cos x}{\sin x}} lalu diakarkan kembali. Misal: \displaystyle a=\sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}}, maka:

\displaystyle a=\sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}}\\ a^2=\frac{\sin x}{\cos x}+2+\frac{\cos x}{\sin x}=\frac{\sin^2 x+2\sin x\cos x+\cos^2 x}{\sin x\cos x}=\frac{(\sin x+\cos x)^2}{\sin x\cos x}\\ a=\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}

Untuk \displaystyle \sqrt{\frac{\sin x}{\cos x}}-\sqrt{\frac{\cos x}{\sin x}}\\ caranya sama. Jadi:

\displaystyle =\frac{1}{2}\int \left ( \sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}} \right )dx+\frac{1}{2}\int \left ( \sqrt{\frac{\sin x}{\cos x}}-\sqrt{\frac{\cos x}{\sin x}} \right )dx\\ =\frac{1}{2}\int \frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}\: dx+\frac{1}{2}\int \frac{\sin x-\cos x}{\sqrt{\sin x\cos x}}\: dx\\ =\frac{1}{2}\int \frac{\sin x+\cos x}{\sqrt{\frac{\sin 2x}{2}}}\: dx+\frac{1}{2}\int \frac{\sin x-\cos x}{\sqrt{\frac{\sin 2x}{2}}}\: dx\\ =\frac{\sqrt{2}}{2}\int \frac{\sin x+\cos x}{\sqrt{\sin 2x}}\: dx+\frac{\sqrt{2}}{2}\int \frac{\sin x-\cos x}{\sqrt{\sin 2x}}\: dx\\ =\frac{1}{\sqrt{2}}\int \frac{\sin x+\cos x}{\sqrt{\sin 2x}}\: dx+\frac{1}{\sqrt{2}}\int \frac{\sin x-\cos x}{\sqrt{\sin 2x}}\: dx\\ =\frac{1}{\sqrt{2}}\int \frac{\sin x+\cos x}{\sqrt{1-(1-\sin 2x)}}\: dx+\frac{1}{\sqrt{2}}\int \frac{\sin x-\cos x}{\sqrt{(1+\sin 2x)-1}}\: dx\\ =\frac{1}{\sqrt{2}}\int \frac{\sin x+\cos x}{\sqrt{1-(\sin^2 x+\cos^2 x-2\sin x\cos x)}}\: dx+\frac{1}{\sqrt{2}}\int \frac{\sin x-\cos x}{\sqrt{(\sin^2 x+\cos^2 x+2\sin x\cos x)-1}}\: dx\\ =\frac{1}{\sqrt{2}}\int \frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x)^2}}\: dx+\frac{1}{\sqrt{2}}\int \frac{\sin x-\cos x}{\sqrt{(\sin x+\cos x)^2-1}}\: dx

Akhirnya bentuk yang diperoleh bisa diselesaikan dengan integral substiusi. Untuk \displaystyle \frac{1}{\sqrt{2}}\int \frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x)^2}}\: dx

\displaystyle u=\sin x-\cos x\rightarrow \frac{du}{dx}=\cos x+\sin x\\ =\frac{1}{\sqrt{2}}\int \frac{\sin x+\cos x}{\sqrt{1-u^2}}\: \frac{du}{\cos x+\sin x}\\ =\frac{1}{\sqrt{2}}\arcsin (u)=\frac{\sin^{-1}(\sin x-\cos x)}{\sqrt{2}}

dan untuk \displaystyle \frac{1}{\sqrt{2}}\int \frac{\sin x-\cos x}{\sqrt{(\sin x+\cos x)^2-1}}\: dx

\displaystyle v=\sin x+\cos x\rightarrow \frac{dv}{dx}=\cos x-\sin x\\ =\frac{1}{\sqrt{2}}\int \frac{\sin x-\cos x}{\sqrt{v^2-1}}\frac{dv}{\cos x-\sin x}\\ =-\frac{1}{\sqrt{2}}\int \frac{\cos x-\sin x}{\sqrt{v^2-1}}\frac{dv}{\cos x-\sin x}\\ =-\frac{1}{\sqrt{2}}\: \mathrm{arcosh}(v)=-\frac{\cosh^{-1}(\sin x+\cos x)}{\sqrt{2}}

Hasil dari pengintegralan ∫ √tan x dx adalah:

\displaystyle \int \sqrt{\tan x}\: dx\\ =\frac{1}{\sqrt{2}}\int \frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x)^2}}\: dx+\frac{1}{\sqrt{2}}\int \frac{\sin x+\cos x}{\sqrt{(\sin x+\cos x)^2-1}}\: dx\\ =\frac{\sin^{-1}(\sin x-\cos x)}{\sqrt{2}}+\left [ -\frac{\cosh^{-1}(\sin x+\cos x)}{\sqrt{2}} \right ]+C\\ =\frac{\sin^{-1}(\sin x-\cos x)-\cosh^{-1}(\sin x+\cos x)}{\sqrt{2}}+C

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